UP Board Solutions Class 11 Maths Chapter 4 Principle of Mathematical Induction

प्रश्नावली 4.1

Question 1. 1 + 3 + 3\(^2\) + ..... + 3\(^{n-1}\) = \( \frac{3^n-1}{2} \)
Answer: हल : माना दिया हुआ कथन P(n) है।
\( \therefore \) P(n): 1 + 3 + 3\(^2\) +.... + 3\(^{n-1}\) = \( \frac{3^n-1}{2} \)
n = 1 रखने पर, \( \therefore \) बायाँ पक्ष P(n) = 1
दायाँ पक्ष = \( \frac{3^1-1}{2} \)
= \( \frac{3-1}{2} \)
= \( \frac{2}{2} \) = 1
\( \therefore \) n = 1 के लिए P(n) सत्य है।
मान लीजिए n = k के लिए कथन सत्य है।
\( \therefore \) P(k) = 1 + 3 + 3\(^2\) + .... + 3\(^{k-1}\) = \( \frac{3^k-1}{2} \)
3\(^k\) को दोनों पक्षों में जोड़ने पर,
1 + 3 + 3\(^2\) +..... + 3\(^{k-1}\) + 3\(^k\) = \( \frac{3^k-1}{2} \) + 3\(^k\)
दायाँ पक्ष = \( \frac{3^k-1+2.3^k}{2} \)
= \( \frac{3^k(1+2)-1}{2} \)
= \( \frac{3.3^k-1}{2} \)
= \( \frac{3^{k+1}-1}{2} \)
\( \therefore \) 1 + 3 + 3\(^2\) +..... + 3\(^k\) = \( \frac{3^{k+1}-1}{2} \)
यह कथन n = k + 1 के लिए सत्य है।
\( \implies \) जब भी P(k) सत्य होगा P(k+1) भी सत्य है।
अंतः गणितीय आगमन सिद्धातं के अनुसार P(n) उन सभी n के मान के लिए सत्य है जो n \( \in \) N है।
In simple words: This solution uses mathematical induction to prove the given series formula. It establishes the base case for n=1, assumes the formula holds for n=k, and then proves it holds for n=k+1 by adding the (k+1)th term to both sides.

🎯 Exam Tip: Clearly show the base case (n=1), the inductive hypothesis (assuming P(k) is true), and the inductive step (proving P(k+1) is true using P(k)) for full marks in mathematical induction problems.

Question 2. 1\(^3\) + 2\(^3\) + 3\(^3\) +...... + n\(^3\) = \( \left( \frac{n(n+1)}{2} \right)^2 \)
Answer: हल : माना
P(n): 1\(^3\) + 2\(^3\) + 3\(^3\) +.....+ n\(^3\) = \( \frac{n^2(n+1)^2}{4} \)
यदि n = 1, बायाँ पक्ष = 1\(^3\) = 1
दायाँ पक्ष = \( \frac{1^2(1+1)^2}{4} \)
= \( \frac{1.4}{4} \) = 1
\( \therefore \) n = 1 के लिए P(n) सत्य है।
मान लीजिए कथन n = k के लिए सत्य है।
\( \therefore \) 1\(^3\) + 2\(^3\) + 3\(^3\) +....+ k\(^3\) = \( \frac{k^2(k+1)^2}{4} \)
दोनों और (k+1)\(^3\) जोड़ने पर,
1\(^3\) + 2\(^3\) + 3\(^3\) +....+ k\(^3\) + (k + 1)\(^3\) = \( \frac{k^2(k+1)^2}{4} \) + (k + 1)\(^3\)
= (k + 1)\(^2\) \( \left[ \frac{k^2}{4} + (k+1) \right] \)
= (k + 1)\(^2\) \( \left[ \frac{k^2+4k+4}{4} \right] \)
= \( \frac{(k+1)^2 (k+2)^2}{4} \)
= \( \frac{(k+1)^2 ( (k+1)+1)^2}{4} \)
इससे सिद्ध हुआ कि यदि P(n) मान n = k के लिए सत्य है तो P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n के सभी मान के लिए सत्य होगा यदि n \( \in \) N.
In simple words: This problem proves the formula for the sum of cubes using mathematical induction. It verifies the formula for n=1, assumes it holds for n=k, and then demonstrates its validity for n=k+1 by adding the (k+1)th cubic term to both sides.

🎯 Exam Tip: When dealing with sums of powers, remember to factor out common terms like \((k+1)^2\) to simplify the expression efficiently during the inductive step. This is a common pattern in these types of proofs.

Question 3. 1 + \( \frac{1}{1+2} \) + \( \frac{1}{1+2+3} \) +.......+ \( \frac{1}{1+2+3+.....+n} \) = \( \frac{2n}{n+1} \)
Answer: हल : माना
P(n) = 1 + \( \frac{1}{1+2} \) + \( \frac{1}{1+2+3} \) +.....+ \( \frac{1}{1+2+3+.....+n} \) = \( \frac{2n}{n+1} \)
n = 1 के लिए बायाँ पक्ष = 1
दायाँ पक्ष = \( \frac{2 \times 1}{1+1} \) = \( \frac{2}{2} \) = 1 = बायाँ पक्ष
\( \therefore \) P(n) n = 1 के लिए सत्य है
मान लिया n = k के लिए कथन सत्य है।
\( \therefore \) P(k) = 1 + \( \frac{1}{1+2} \) + \( \frac{1}{1+2+3} \) +....+ \( \frac{1}{1+2+3+....+k} \) = \( \frac{2k}{k+1} \)
दोनों पक्षों में \( \frac{1}{1+2+3+....+k+(k+1)} \) जोड़ने पर,
1 + \( \frac{1}{1+2} \) + \( \frac{1}{1+2+3} \) +.....+ \( \frac{1}{1+2+3+...+k} \) + \( \frac{1}{1+2+3+...+k+(k+1)} \)
दायाँ पक्ष = \( \frac{2k}{k+1} \) + \( \frac{1}{(k+1)(k+2)/2} \) [: 1 + 2 +....+ n = \( \frac{n(n+1)}{2} \) ]
= \( \frac{2k}{k+1} \) + \( \frac{2}{(k+1)(k+2)} \)
= \( \frac{2k(k+2) + 2}{(k+1)(k+2)} \)
= \( \frac{2k^2 + 4k + 2}{(k+1)(k+2)} \)
= \( \frac{2[k^2 + 2k + 1]}{(k+1)(k+2)} \)
= \( \frac{2(k+1)^2}{(k+1)(k+2)} \)
= \( \frac{2(k+1)}{k+2} \)
= \( \frac{2(k+1)}{k+1+1} \)
इससे सिद्ध हुआ कि P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This proof uses mathematical induction to confirm a formula for a series of sums. It establishes the base case for n=1, assumes the formula holds for n=k, and then shows it holds for n=k+1 by adding the next term and simplifying the expression.

🎯 Exam Tip: When working with series, correctly identifying the formula for the sum of the first 'n' natural numbers (\(\frac{n(n+1)}{2}\)) is crucial. Also, algebra skills, especially factoring and combining fractions, are key in the inductive step.

Question 4. 1.2.3 + 2.3.4 +.....+ n(n + 1)(n + 2) = \( \frac{n(n+1)(n+2)(n+3)}{4} \)
Answer: हल : मान लीजिए
P(n) = 1.2.3 + 2.3.4 +.....+ n(n + 1)(n + 2) = \( \frac{n(n+1)(n+2)(n+3)}{4} \)
यदि n = 1, बायाँ पक्ष = 1.2.3 = 6
दायाँ पक्ष = \( \frac{1.2.3.4}{4} \) = 6
\( \therefore \) n = 1 के लिए P(n) सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
1.2.3 + 2.3.4 + 3.4.5 +....+ k (k + 1)(k + 2) = \( \frac{k(k+1)(k+2)(k+3)}{4} \)
पक्ष का (k + 1)\(^{th}\) पद = (k + 1)(k + 2)(k + 3)
दोनों पक्षों में (k + 1)(k + 2)(k + 3) जोड़ने पर
1.2.3 + 2.3.4 + 3.4.5 +....+ k (k + 1)(k + 2) + (k + 1)(k + 2)(k + 3)
= \( \frac{k(k+1)(k+2)(k+3)}{4} \) + (k + 1)(k + 2)(k + 3)
= (k + 1)(k + 2)(k + 3) \( \left[ \frac{k}{4}+1 \right] \)
= (k + 1)(k + 2)(k + 3) \( \left[ \frac{k+4}{4} \right] \)
= \( \frac{(k+1)(k+2)(k+3)(k+4)}{4} \)
= \( \frac{(k+1)((k+1)+1)((k+1)+2)((k+1)+3)}{4} \)
इससे सिद्ध हुआ कि P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a formula for the sum of products of three consecutive integers using mathematical induction. It involves verifying the base case for n=1, assuming the formula holds for n=k, and then showing it holds for n=k+1 by adding the next product term and simplifying.

🎯 Exam Tip: Look for common factors to extract and simplify algebraic expressions effectively, especially in the inductive step. This reduces complexity and helps in reaching the desired form for P(k+1).

Question 5. 1.3 + 2.3\(^2\) + 3.3\(^3\) +....+ n. 3\(^n\) = \( \frac{(2n-1).3^{n+1}+3}{4} \)
Answer: हल : माना
P(n): 1.3 + 2.3\(^2\) + 3.3\(^3\) +....+ n.3\(^n\) = \( \frac{(2n-1).3^{n+1}+3}{4} \)
यदि n = 1, P(n) का बायाँ पक्ष = 1.3 = 3
दायाँ पक्ष = \( \frac{(2.1-1).3^{1+1}+3}{4} \)
= \( \frac{(1).3^2+3}{4} \)
= \( \frac{1.9+3}{4} \) = \( \frac{12}{4} \) = 3
\( \therefore \) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\( \therefore \) 1.3 + 2.3\(^2\) + 3.3\(^3\) +......+ k.3\(^k\) = \( \frac{(2k-1).3^{k+1}+3}{4} \)
(k+1) वाँ पद = (k + 1).3\(^{k+1}\)
(k + 1).3\(^{k+1}\) को दोनों पक्षों में जोड़ने पर,
1.3 + 2.3\(^2\) + 3.3\(^3\) +...+ k.3\(^k\) + (k + 1).3\(^{k+1}\)
= \( \frac{(2k-1).3^{k+1}+3}{4} \) + (k + 1).3\(^{k+1}\)
दायाँ पक्ष = \( \frac{(2k-1).3^{k+1}+3+4(k+1).3^{k+1}}{4} \)
= \( \frac{3^{k+1}(2k-1+4(k+1))+3}{4} \)
= \( \frac{3^{k+1}(2k-1+4k+4)+3}{4} \)
= \( \frac{3^{k+1}(6k+3)+3}{4} \)
= \( \frac{3.3^{k+1}(2k+1)+3}{4} \)
= \( \frac{3^{k+2}(2k+1)+3}{4} \)
= \( \frac{(2(k+1)-1).3^{k+1+1}+3}{4} \)
इससे सिद्ध हुआ कि P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार, P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem utilizes mathematical induction to prove the given series formula involving terms of n multiplied by powers of 3. The proof establishes the base case for n=1, assumes the formula holds for n=k, and then extends it to n=k+1 by adding the next term and simplifying the expression.

🎯 Exam Tip: Pay close attention to exponent manipulation and factoring out common exponential terms (like \(3^{k+1}\)) to simplify the expression efficiently. Mistakes in these steps are common pitfalls in such proofs.

Question 6. 1.2 + 2.3 + 3.4+...+ n(n + 1) = \( \frac{n(n+1)(n+2)}{3} \)
Answer: हल : माना
P(n) = 1.2 + 2.3 + 3.4 +....+ n(n + 1) = \( \frac{n(n+1)(n+2)}{3} \)
यदि n = 1, बायाँ पक्ष = 1.2 = 2
दायाँ पक्ष = \( \frac{1(1+1)(1+2)}{3} \) = \( \frac{1.2.3}{3} \) = 2
\( \therefore \) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है
\( \therefore \) 1.2 + 2.3 + 3.4+....+ k(k + 1) = \( \frac{k(k+1)(k+2)}{3} \)
(k + 1) वाँ पद = (k + 1)(k + 2)
दोनों पक्षों में (k + 1)(k + 2) जोड़ने पर,
1.2 + 2.3 + 3.4 +....+ k(k + 1) + (k + 1)(k + 2)
= \( \frac{k(k+1)(k+2)}{3} \) + (k + 1)(k + 2)
= (k + 1)(k + 2) \( \left( \frac{k}{3}+1 \right) \)
= (k + 1)(k + 2) \( \left( \frac{k+3}{3} \right) \)
= \( \frac{(k+1)(k+2)(k+3)}{3} \)
= \( \frac{(k+1)((k+1)+1)((k+1)+2)}{3} \)
\( \therefore \) P(n), n = k +1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a formula for the sum of products of two consecutive integers using mathematical induction. The solution verifies the base case for n=1, assumes the formula holds for n=k, and then demonstrates its validity for n=k+1 by adding the next product term and simplifying.

🎯 Exam Tip: Factoring out common terms like \((k+1)(k+2)\) is key to simplifying the expression in the inductive step. This helps in transforming the P(k) expression into the P(k+1) form efficiently.

Question 7. 1.3 + 3.5+ 5.7 +....+ (2n - 1)(2n + 1) = \( \frac{n(4n^2+6n-1)}{3} \)
Answer: हल : मान लीजिए
P(n): 1.3 + 3.5+ 5.7 +....+ (2n - 1)(2n + 1) = \( \frac{n(4n^2+6n-1)}{3} \)
यदि n = 1, बायाँ पक्ष = 1.3 = 3
दायाँ पक्ष = \( \frac{1(4.1^2+6.1-1)}{3} \)
= \( \frac{4+6-1}{3} \) = \( \frac{9}{3} \) = 3
\( \therefore \) P(n), n = 1 के लिए सत्य है।
\( \therefore \) 1.3 + 3.5+ 5.7 +....+ (2k-1)(2k + 1) = \( \frac{k(4k^2+6k-1)}{3} \)
(k + 1) वाँ पद = [2 (k + 1) -1] [2(k + 1) + 1] = (2k+1)(2k+ 3) को दोनों पक्षों में जोड़ने पर,
1.3 + 3.5 + 5.7 +....+(2k-1)(2k+1) + (2k+1)(2k + 3)
= \( \frac{k(4k^2+6k-1)}{3} \) +(2k+1) (2k+ 3)
= \( \frac{k(4k^2+6k-1)+3(2k+1)(2k+3)}{3} \)
= \( \frac{4k^3+6k^2-k+3(4k^2 + 8k +3)}{3} \)
= \( \frac{4k^3+6k^2-k+12k^2+24k+9}{3} \)
= \( \frac{4k^3 +18k^2+23k +9}{3} \)
यहाँ, हम \( (k+1)(4(k+1)^2+6(k+1)-1) \) को विस्तार करके जांचते हैं:
\( (k+1)[4(k^2+2k+1)+6k+6-1] \)
\( (k+1)[4k^2+8k+4+6k+5] \)
\( (k+1)[4k^2+14k+9] \)
\( 4k^3+14k^2+9k+4k^2+14k+9 \)
\( 4k^3+18k^2+23k+9 \)
यह समान है।
\( \implies \) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a formula for a series where each term is a product of two consecutive odd numbers using mathematical induction. The solution verifies the base case for n=1, assumes the formula holds for n=k, and then confirms its validity for n=k+1 by adding the next term and simplifying the resulting expression.

🎯 Exam Tip: The algebraic simplification in the inductive step can be complex. Carefully expand and combine like terms, and then try to factor the resulting polynomial to match the P(k+1) form. Double-checking your polynomial multiplication is essential.

Question 8. 1.2 + 2.2\(^2\) + 3.2\(^3\) +.....+ n.2\(^n\) = (n - 1). 2\(^{n+1}\) + 2.
Answer: हल : माना
P(n): 1.2 + 2.2\(^2\) + 3.2\(^3\) +....+ n.2\(^n\) = (n - 1). 2\(^{n+1}\) + 2
यदि n = 1, बायाँ पक्ष = 1.2 = 2
दायाँ पक्ष = (1 - 1). 2\(^{1+1}\) + 2 = 0.2\(^2\) + 2 = 0 + 2 = 2
\( \therefore \) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\( \therefore \) 1.2 + 2.2\(^2\) + 3.2\(^3\) +....+ k.2\(^k\) = (k - 1).2\(^{k+1}\) + 2
(k + 1) वाँ पद = (k + 1).2\(^{k+1}\) को दोनों पक्षों में जोड़ने पर,
1.2 + 2.2\(^2\) + 3.2\(^3\) +.....+ k.2\(^k\) + (k + 1).2\(^{k+1}\) = (k - 1).2\(^{k+1}\) +2 + (k + 1).2\(^{k+1}\)
= 2\(^{k+1}\) (k - 1 + k + 1) + 2
= 2\(^{k+1}\). (2k) + 2
= 2.k.2\(^{k+1}\) + 2 = k.2\(^{k+2}\) + 2
हमें इसे \( ((k+1)-1).2^{((k+1)+1)} + 2 \) के रूप में दिखाना है
\( k.2^{k+2} + 2 = k.2^{k+1+1} + 2 \)
यह \( P(k+1) \) का स्वरूप है।
\( \implies \) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार, P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This solution uses mathematical induction to prove a formula for a series where each term is the product of an integer and a power of 2. It confirms the base case, assumes the formula holds for k, and then shows it holds for k+1 by adding the next term and simplifying the expression using exponent rules.

🎯 Exam Tip: Be careful with the algebraic manipulation of exponents and factoring, especially when combining terms like \((k-1).2^{k+1}\) and \((k+1).2^{k+1}\). The goal is to isolate \(2^{k+1}\) or \(2^{k+2}\) and simplify the coefficients.

Question 9. \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.....+ \frac{1}{2^n} = 1 - \frac{1}{2^n} \)
Answer: हल : माना P(n): \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.....+ \frac{1}{2^n} = 1 - \frac{1}{2^n} \)
यदि n = 1, बायाँ पक्ष = \( \frac{1}{2} \)
दायाँ पक्ष = 1 - \( \frac{1}{2^1} \) = 1 - \( \frac{1}{2} \) = \( \frac{1}{2} \)
\( \therefore \) P(n), n = 1 के लिए सत्य है।
मान लिया P(n), n = k के लिए सत्य है।
\( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.....+ \frac{1}{2^k} = 1 - \frac{1}{2^k} \)
(k + 1) वाँ पद = \( \frac{1}{2^{k+1}} \) दोनों पक्षों में जोड़ने पर
\( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +.....+ \frac{1}{2^k} + \frac{1}{2^{k+1}} = 1 - \frac{1}{2^k} + \frac{1}{2^{k+1}} \)
= 1 - \( \left( \frac{1}{2^k} - \frac{1}{2^{k+1}} \right) \)
= 1 - \( \frac{2-1}{2^{k+1}} \)
= 1 - \( \frac{1}{2^{k+1}} \)
\( \implies \) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves the formula for the sum of a geometric progression (series of powers of 1/2) using mathematical induction. It verifies the base case for n=1, assumes the formula holds for n=k, and then shows its validity for n=k+1 by adding the next term and simplifying the expression.

🎯 Exam Tip: When simplifying fractions with exponents, find a common denominator that is the highest power of the base. For example, when combining \(-\frac{1}{2^k} + \frac{1}{2^{k+1}}\), use \(2^{k+1}\) as the common denominator to efficiently reach the desired form.

Question 10. \( \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} +....+ \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
Answer: हल : माना
P(n) : \( \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} +....+ \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4} \)
यदि n = 1, बायाँ पक्ष = \( \frac{1}{2.5} \) = \( \frac{1}{10} \)
दायाँ पक्ष = \( \frac{1}{6.1+4} \) = \( \frac{1}{10} \)
\( \implies \) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\( \therefore \) \( \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} +....+ \frac{1}{(3k-1)(3k+2)} = \frac{k}{6k+4} \)
(k + 1) वाँ पद = \( \frac{1}{[3(k+1)-1][3(k+1)+2]} \) = \( \frac{1}{(3k+2)(3k+5)} \) दोनों पक्षों में जोड़ने पर
\( \frac{1}{2.5} + \frac{1}{5.8} + \frac{1}{8.11} +....+ \frac{1}{(3k-1)(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
= \( \frac{k}{6k+4} + \frac{1}{(3k+2)(3k+5)} \)
= \( \frac{k}{2(3k+2)} + \frac{1}{(3k+2)(3k+5)} \)
= \( \frac{k(3k+5)+2}{2(3k+2)(3k+5)} \)
= \( \frac{3k^2 + 5k +2}{2(3k+2)(3k+5)} \)
= \( \frac{(3k+2)(k+1)}{2(3k+2)(3k+5)} \)
= \( \frac{k+1}{2(3k+5)} \)
= \( \frac{k+1}{6k+10} \)
= \( \frac{k+1}{6(k+1)+4} \)
\( \implies \) P(n), n = k+ 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a formula for a series of fractions where the denominator is a product of two terms from an arithmetic progression. The solution uses mathematical induction, verifying the base case for n=1, assuming the formula holds for n=k, and then showing it holds for n=k+1 by adding the next fractional term and simplifying the expression.

🎯 Exam Tip: When dealing with fractional series, partial fraction decomposition (though not explicitly used here, the form suggests it) or finding a common denominator and simplifying the numerator is crucial. Factor the quadratic in the numerator to identify common factors for simplification.

Question 11. \( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} +....+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \)
Answer: हल : माना
P(n): \( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} +....+ \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)} \)
n = 1 के लिए बायाँ पक्ष = \( \frac{1}{1.2.3} \) = \( \frac{1}{6} \)
दायाँ पक्ष = \( \frac{1(1+3)}{4(1+1)(1+2)} \)
= \( \frac{1.4}{4.2.3} \) = \( \frac{4}{24} \) = \( \frac{1}{6} \)
\( \implies \) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\( \therefore \) \( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} +....+ \frac{1}{k(k+1)(k+2)} = \frac{k(k+3)}{4(k+1)(k+2)} \)
(k + 1) वाँ पद = \( \frac{1}{(k+1)(k+2)(k+3)} \) दोनों पक्षों में जोड़ने पर,
\( \frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} +....+ \frac{1}{k(k+1)(k+2)} + \frac{1}{(k+1)(k+2)(k+3)} \)
= \( \frac{k(k+3)}{4(k+1)(k+2)} \) + \( \frac{1}{(k+1)(k+2)(k+3)} \)
= \( \frac{1}{(k+1)(k+2)} \) \( \left[ \frac{k(k+3)}{4} + \frac{1}{k+3} \right] \) - Error in OCR/Math equation here. It should be \( \frac{1}{(k+1)(k+2)} \) \( \left[ \frac{k(k+3)}{4} + \frac{1}{(k+3)} \right] \) is incorrect.
Corrected step:
= \( \frac{1}{(k+1)(k+2)} \) \( \left[ \frac{k(k+3)}{4} + \frac{1}{k+3} \right] \) is incorrect.
The correct common denominator would be \(4(k+1)(k+2)(k+3)\).
Let's re-evaluate:
= \( \frac{k(k+3)(k+3) + 4}{4(k+1)(k+2)(k+3)} \)
= \( \frac{k(k^2+6k+9) + 4}{4(k+1)(k+2)(k+3)} \)
= \( \frac{k^3+6k^2+9k + 4}{4(k+1)(k+2)(k+3)} \)
Now we need to factor the numerator \( k^3+6k^2+9k+4 \). By rational root theorem, possible roots are \(\pm 1, \pm 2, \pm 4\).
If \( k=-1 \), \( (-1)^3+6(-1)^2+9(-1)+4 = -1+6-9+4 = 0 \). So \((k+1)\) is a factor.
Dividing \( k^3+6k^2+9k+4 \) by \((k+1)\) gives \( k^2+5k+4 \).
So, \( k^3+6k^2+9k+4 = (k+1)(k^2+5k+4) \).
Further factorizing \( k^2+5k+4 = (k+1)(k+4) \).
Thus, \( k^3+6k^2+9k+4 = (k+1)(k+1)(k+4) = (k+1)^2(k+4) \).
So,
= \( \frac{(k+1)^2(k+4)}{4(k+1)(k+2)(k+3)} \)
= \( \frac{(k+1)(k+4)}{4(k+2)(k+3)} \)
This needs to be in the form \( \frac{(k+1)((k+1)+3)}{4((k+1)+1)((k+1)+2)} \).
Let's check if \( \frac{(k+1)(k+4)}{4(k+2)(k+3)} \) matches this.
Yes, \( k+4 = (k+1)+3 \), \( k+2 = (k+1)+1 \), \( k+3 = (k+1)+2 \).
So it matches.
The OCR had:
= \( \frac{(k+1)(k^2 + 5k + 4)}{4(k+1)(k+2)(k + 3)} \)
= \( \frac{(k+4)(k+1)}{4(k+2)(k+3)} \)
= \( \frac{(k+1)(k+1+3)}{4(k+1+1)(k+1+2)} \)
\( \implies \) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \( \in \) N, n के सभी मानों के लिए सत्य है।
In simple words: This solution applies mathematical induction to prove a formula for a series of fractions with three consecutive terms in the denominator. It confirms the base case, assumes the formula holds for n=k, and then demonstrates its validity for n=k+1 by adding the next term and performing complex algebraic simplification involving polynomial factorization.

🎯 Exam Tip: When dealing with complex rational expressions in the inductive step, factorizing both the numerator and denominator is crucial. Long division or synthetic division can help factor cubic polynomials in the numerator once a root is identified (e.g., using the Rational Root Theorem).

Question 12. a + ar + ar² +.....+ ar^{n - 1} = \(\frac{a(r^n -1)}{r-1}\).
Answer: हल : माना लीजिए P(n) = a + ar + ar² +.....+ ar^n-1 = \(\frac{a(r^n -1)}{r-1}\),
n = 1 के लिए बायाँ पक्ष = a
दायाँ पक्ष = \(\frac{a(r^1 -1)}{r-1}\) = \(\frac{a(r-1)}{r-1}\) = a
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) a + ar + ar² +.....+ ar^k-1 = \(\frac{a(1-r^k)}{1-r}\)
(k + 1) वाँ पद = ar^k को दोनों पक्षों में जोड़ने पर,
a + ar + ar² +......+ ar^k-1 + ar^k = \(\frac{a(1-r^k)}{1-r}\) + ar^k
= a \(\left[\frac{1-r^k}{1-r} + r^k\right]\)
= a \(\left[\frac{1-r^k + r^k(1-r)}{1-r}\right]\)
= a \(\left[\frac{1-r^k + r^k - r^{k+1}}{1-r}\right]\)
= \(\frac{a(1-r^{k+1})}{1-r}\)
= \(\frac{a(r^{k+1}-1)}{r-1}\)
\(\implies\) P(n), n = k +1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a geometric progression sum formula using mathematical induction. We verify the base case for n=1, assume it holds for n=k, and then show it also holds for n=k+1 by adding the (k+1)-th term to both sides and simplifying the expression to match the formula for n=k+1.

🎯 Exam Tip: When proving summation formulas using induction, ensure algebraic simplification is meticulous, especially when combining fractions with different denominators. The final form for P(k+1) must exactly match the original formula with n replaced by (k+1).

Question 13. \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)....\left(1+\frac{2n+1}{n^2}\right)\) = (n + 1)².
Answer: हल : माना
P(n): \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)....\left(1+\frac{2n+1}{n^2}\right)\) = (n + 1)²
n = 1 के लिए बायाँ पक्ष = \(\left(1+\frac{3}{1}\right)\) = 1 + 3 = 4
दायाँ पक्ष = (n + 1)² = (1 + 1)² = 2² = 4
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)....\left(1+\frac{2k+1}{k^2}\right)\) = (k + 1)²
(k + 1) वाँ पद = \(\left(1+\frac{2(k+1)+1}{(k+1)^2}\right)\) = \(\left(1+\frac{2k+2+1}{(k+1)^2}\right)\) = \(\left(1+\frac{2k+3}{(k+1)^2}\right)\) से दोनों पक्षों में गुणा करने पर,
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)....\left(1+\frac{2k+1}{k^2}\right)\left(1+\frac{2k+3}{(k+1)^2}\right)\)
= (k + 1)² \(\left(1+\frac{2k+3}{(k+1)^2}\right)\)
= (k + 1)² \(\left(\frac{(k+1)^2+2k+3}{(k+1)^2}\right)\)
= (k + 1)² + 2k + 3
= k² + 2k + 1 + 2k + 3
= k² + 4k + 4
= (k + 2)²
= \((k+1+1)^2\)
\(\implies\) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses mathematical induction to prove a product series identity. We establish the base case for n=1. Then, assuming the identity holds for n=k, we multiply both sides by the (k+1)-th term of the series. After algebraic simplification, the resulting expression should match the original formula with n replaced by (k+1).

🎯 Exam Tip: For product series induction, carefully expand and factorize algebraic expressions. The goal is to transform the P(k+1) expression into the required (k+2)² form, demonstrating the inductive step successfully.

Question 14. \(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{n}\right)\) = n + 1.
Answer: हल : माना P(n) : \(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{n}\right)\) = n + 1
n = 1 के लिए बायाँ पक्ष = \(\left(1+\frac{1}{1}\right)\) = 1 + 1 = 2
दायाँ पक्ष = n + 1 = 1 + 1 = 2
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) \(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{k}\right)\) = k + 1
(k + 1) वाँ गुणनखंड = \(\left(1+\frac{1}{k+1}\right)\) से दोनों पक्षों में गुणा करने पर,
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)\)
= (k + 1)\(\left(1+\frac{1}{k+1}\right)\)
= (k + 1)\(\left(\frac{k+1+1}{k+1}\right)\)
= k + 1 + 1
= k + 2
= (k+1)+1
इससे सिद्ध होता है कि P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a product series identity by mathematical induction. The base case for n=1 is confirmed. Then, assuming the identity holds for n=k, we multiply both sides by the (k+1)-th term. After straightforward algebraic simplification, the result matches the formula for n=k+1, thereby completing the induction.

🎯 Exam Tip: Simple product series proofs like this often involve cancellations or direct addition/subtraction. Pay close attention to how the (k+1)-th term integrates with the existing P(k) expression.

Question 15. 1² + 3² + 5² +.....+ (2n - 1)² = \(\frac{n(2n-1)(2n+1)}{3}\).
Answer: हल : माना
P(n): 1² + 3² + 5² +.....+ (2n - 1)² = \(\frac{n(2n-1)(2n+1)}{3}\)
n = 1 के लिए बायाँ पक्ष = 1² = 1
दायाँ पक्ष = \(\frac{1(2 \cdot 1-1)(2 \cdot 1+1)}{3}\) = \(\frac{1(1)(3)}{3}\) = 1
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) 1² + 3² + 5² +.....+ (2k - 1)² = \(\frac{k(2k-1)(2k+1)}{3}\)
(k + 1) वाँ पद = (2(k+1) - 1)² = (2k + 2 - 1)² = (2k + 1)² को दोनों पक्षों में जोड़ने पर,
1² + 3² + 5² +.....+ (2k - 1)² + (2k + 1)²
= \(\frac{k(2k-1)(2k+1)}{3}\) + (2k + 1)²
= (2k + 1) \(\left[\frac{k(2k-1)}{3} + (2k + 1)\right]\)
= (2k + 1) \(\left[\frac{k(2k-1)+3(2k+1)}{3}\right]\)
= (2k + 1) \(\left[\frac{2k^2-k+6k+3}{3}\right]\)
= (2k + 1) \(\left[\frac{2k^2+5k+3}{3}\right]\)
= \(\frac{(2k+1)(2k+3)(k+1)}{3}\)
= \(\frac{(k+1)[2(k+1)-1][2(k+1)+1]}{3}\)
\(\implies\) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses mathematical induction to prove the sum of squares of odd numbers. We first confirm the base case for n=1. Then, assuming the formula holds for n=k, we add the (k+1)-th odd square term to both sides. The key is to factor out the common term (2k+1) and simplify the remaining quadratic expression, eventually matching the form for n=k+1.

🎯 Exam Tip: Factoring common terms and simplifying quadratic expressions are crucial. Ensure that the numerator factors correctly into terms resembling (2(k+1)-1) and (2(k+1)+1) along with (k+1) to reach the target form.

Question 16. \(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} +.....+\frac{1}{(3n-2)(3n+1)}\) = \(\frac{n}{3n+1}\).
Answer: हल : माना
P(n): \(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} +.....+\frac{1}{(3n-2)(3n+1)}\) = \(\frac{n}{3n+1}\)
n = 1 के लिए बायाँ पक्ष = \(\frac{1}{1 \cdot 4}\) = \(\frac{1}{4}\)
दायाँ पक्ष = \(\frac{1}{3 \cdot 1+1}\) = \(\frac{1}{4}\)
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) \(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} +.....+\frac{1}{(3k-2)(3k+1)}\) = \(\frac{k}{3k+1}\)
(k + 1) वाँ पद = \(\frac{1}{(3(k+1)-2)(3(k+1)+1)}\) = \(\frac{1}{(3k+3-2)(3k+3+1)}\) = \(\frac{1}{(3k+1)(3k+4)}\) को दोनों पक्षों में जोड़ने पर,
\(\frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} +.....+\frac{1}{(3k-2)(3k+1)}\) + \(\frac{1}{(3k+1)(3k+4)}\)
= \(\frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)}\)
= \(\frac{1}{3k+1} \left[ k + \frac{1}{3k+4} \right]\)
= \(\frac{1}{3k+1} \left[ \frac{k(3k+4)+1}{3k+4} \right]\)
= \(\frac{3k^2+4k+1}{(3k+1)(3k+4)}\)
= \(\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}\)
= \(\frac{k+1}{3k+4}\)
= \(\frac{k+1}{3(k+1)+1}\)
\(\implies\) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves a series sum involving fractions using mathematical induction. After verifying the base case for n=1, we assume the formula is true for n=k. We then add the (k+1)-th term to both sides, find a common denominator, and simplify. The simplification involves factoring the quadratic numerator to match the form for n=k+1.

🎯 Exam Tip: When dealing with sums of fractions, correctly finding a common denominator and factoring the numerator are critical steps. Look for factors that relate to (k+1) and the structure of the denominator in the original formula.

Question 17. \(\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} +.....+\frac{1}{(2n+1)(2n+3)}\) = \(\frac{n}{3(2n+3)}\).
Answer: हल : माना
P(n): \(\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} +.....+\frac{1}{(2n+1)(2n+3)}\) = \(\frac{n}{3(2n+3)}\)
n = 1 के लिए बायाँ पक्ष = \(\frac{1}{3 \cdot 5}\) = \(\frac{1}{15}\)
दायाँ पक्ष = \(\frac{1}{3(2 \cdot 1+3)}\) = \(\frac{1}{3(5)}\) = \(\frac{1}{15}\)
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लिया P(n), n = k के लिए सत्य है।
\(\therefore\) \(\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} +.....+\frac{1}{(2k+1)(2k+3)}\) = \(\frac{k}{3(2k+3)}\)
(k + 1) वाँ पद = \(\frac{1}{(2(k+1)+1)(2(k+1)+3)}\) = \(\frac{1}{(2k+3)(2k+5)}\) को दोनों पक्षों में जोड़ने पर,
\(\frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \frac{1}{7 \cdot 9} +.....+\frac{1}{(2k+1)(2k+3)}\) + \(\frac{1}{(2k+3)(2k+5)}\)
= \(\frac{k}{3(2k+3)} + \frac{1}{(2k+3)(2k+5)}\)
= \(\frac{1}{2k+3} \left[ \frac{k}{3} + \frac{1}{2k+5} \right]\)
= \(\frac{1}{2k+3} \left[ \frac{k(2k+5)+3}{3(2k+5)} \right]\)
= \(\frac{2k^2+5k+3}{3(2k+3)(2k+5)}\)
= \(\frac{(k+1)(2k+3)}{3(2k+3)(2k+5)}\)
= \(\frac{k+1}{3(2k+5)}\)
= \(\frac{k+1}{3(2(k+1)+3)}\)
\(\implies\) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem involves proving a fractional series sum using mathematical induction. After confirming the base case for n=1, we assume the formula holds for n=k. We then add the (k+1)-th term to both sides, combine the fractions using a common denominator, and factor the resulting quadratic in the numerator to show it matches the formula for n=k+1.

🎯 Exam Tip: Pay careful attention to the factorization of the numerator and cancellation of terms. The goal is to isolate the (k+1) factor and the denominator structure required for P(k+1).

Question 18. 1 + 2 + 3 +......+ n < \(\frac{1}{8}\)(2n + 1)².
Answer: हल : माना P(n) : 1 + 2 + 3 +.....+ n < \(\frac{1}{8}\)(2n + 1)²
n = 1 के लिए बायाँ पक्ष = 1
दायाँ पक्ष = \(\frac{1}{8}\)(2 \(\cdot\) 1 + 1)² = \(\frac{1}{8}\)(3)² = \(\frac{9}{8}\)
1 < \(\frac{9}{8}\)
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) 1 + 2 + 3 +.....+ k < \(\frac{1}{8}\)(2k + 1)²
(k + 1) वाँ पद = k + 1 दोनों पक्षों में जोड़ने पर
बायाँ पक्ष = 1 + 2 + 3 +.....+ k + (k + 1)
< \(\frac{1}{8}\)(2k + 1)² + (k + 1)
= \(\frac{1}{8}\)[(2k + 1)² + 8(k + 1)]
= \(\frac{1}{8}\)[4k² + 4k + 1 + 8k + 8]
= \(\frac{1}{8}\)[4k² + 12k + 9]
= \(\frac{1}{8}\)(2k + 3)²
= \(\frac{1}{8}\)(2(k + 1) + 1)²
\(\implies\) P(n), n = k + 1 के लिए सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves an inequality using mathematical induction. We first verify the inequality for n=1. Assuming it holds for n=k, we add (k+1) to both sides of the inequality. Then, by simplifying the right-hand side, we show that it becomes less than \(\frac{1}{8}\)(2(k+1)+1)², thus completing the inductive step.

🎯 Exam Tip: For inequalities, algebraic manipulation needs to preserve the inequality direction. Pay special attention to squaring binomials and combining terms to reach the exact form for P(k+1) while maintaining the '<' sign.

Question 19. n(n + 1)(n + 5), संख्या 3 का एक गुणज है।
Answer: हल : मान लीजिए P(n) : n(n+1)(n + 5), संख्या 3 का गुणज है
n = 1 के लिए n(n + 1)(n + 5) = 1 \(\cdot\) 2 \(\cdot\) 6 = 12 जो 3 का गुणज है
P(n), n = 1 के लिए सत्य है
मान लीजिए P(n), n = k के लिए सत्य है
k(k + 1)(k + 5) = 3m
या k³ + 6k² + 5k = 3m
k के स्थान पर k + 1 रखने पर
(k + 1)³ + 6(k + 1)² + 5(k + 1)
= (k³ + 3k² + 3k + 1) + 6(k² + 2k + 1) + 5k + 5
= k³ + 3k² + 3k + 1 + 6k² + 12k + 6 + 5k + 5
= k³ + 9k² + 20k + 12
= (k³ + 6k² + 5k) + (3k² + 15k + 12)
= 3m + 3(k² + 5k + 4)
यह 3 का एक गुणज है।
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses induction to show that the product n(n+1)(n+5) is always divisible by 3. We confirm the base case for n=1. Assuming it's true for n=k, we write k(k+1)(k+5) as 3m. Then, we substitute (k+1) into the expression, expand, and rearrange terms to show it can be expressed as 3m plus another multiple of 3, thus proving divisibility by 3 for n=k+1.

🎯 Exam Tip: When proving divisibility, expand the P(k+1) expression and strategically group terms to reintroduce the P(k) assumption (3m). The remaining terms must also be shown to be a multiple of the divisor (3 in this case).

Question 20. 10^2n-1 + 1, संख्या 11 से भाज्य है।
Answer: हल : माना P(n) : 10^{2n - 1} + 1 संख्या 11 से विभाजित होती है।
n = 1, के लिए 10^{2 \(\cdot\) 1 - 1} + 1 = 10¹ + 1 = 11
P(n), n = 1 के लिए सत्य है
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) 10^2k-1+1, संख्या 11 से विभाजित होती है।
या 10^2k-1+1 = 11m (माना)
k को k + 1 से बदलने पर
10^2(k+1)-1 + 1 = 10^2k+2-1 + 1
= 10^2k+1 + 1
= 10² \(\cdot\) 10^2k-1 + 1
= 100 \(\cdot\) 10^2k-1 + 1
अब, 10^2k-1 = 11m - 1 रखने पर
= 100 \(\cdot\) (11m - 1) + 1
= 100 \(\cdot\) 11m - 100 + 1
= 100 \(\cdot\) 11m - 99
= 11 (100m - 9)
इससे सिद्ध हुआ कि 10^2k+1 + 1 भी 11 से विभाजित होता है।
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses induction to show that \(10^{2n-1} + 1\) is divisible by 11. We verify the base case for n=1 (11 is divisible by 11). Assuming it's true for n=k, we write \(10^{2k-1} = 11m - 1\). Then, we substitute this into the expression for n=k+1, rearrange, and factor out 11 to prove divisibility.

🎯 Exam Tip: For divisibility proofs, the key is to strategically substitute the P(k) assumption into the P(k+1) expression and manipulate it algebraically to clearly show the desired factor (11 in this case).

Question 21. x²ⁿ - y²ⁿ, (x + y) से भाज्य है।
Answer: हल : मान लीजिए P(n): x²ⁿ-y²ⁿ, x + y से विभाजित होता है।
n = 1 के लिए x² - y² = (x - y)(x + y) जो x + y से विभक्त होता है।
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) x^2k-y^2k, x + y से विभक्त होता है।
या x^2k-y^2k = m(x + y)
या x^2k = m(x + y) + y^2k ...(1)
k के स्थान पर k + 1 रखने पर, सिद्ध करना है कि x^2(k+1) - y^2(k+1), x + y से विभक्त होता है।
x^2(k+1)-y^2(k+1) = x^2k+2 - y^2k+2
= x² \(\cdot\) x^2k - y^2k+2
(1) से x^2k का मान रखने पर,
= x² [m(x + y) + y^2k] - y^2k+2
= m(x + y)x² + x²y^2k - y^2k+2
= m(x + y)x² + y^2k(x² - y²)
= m(x + y)x² + y^2k(x - y)(x + y)
= (x + y) [mx² + y^2k(x - y)]
इससे सिद्ध होता है कि x^2(k+1) - y^2(k+1), x + y से विभाजित होता है।
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses mathematical induction to prove that \(x^{2n} - y^{2n}\) is divisible by \(x+y\). We verify the base case for n=1, where \(x^2 - y^2 = (x-y)(x+y)\), which is clearly divisible by \(x+y\). Assuming it's true for n=k, we express \(x^{2k} - y^{2k}\) as a multiple of \(x+y\). Then, for n=k+1, we manipulate \(x^{2(k+1)} - y^{2(k+1)}\) by substituting the P(k) assumption and factoring to show it is also a multiple of \(x+y\).

🎯 Exam Tip: For divisibility proofs involving algebraic expressions, strategically replace terms using the P(k) assumption. Factoring out the divisor \((x+y)\) is the ultimate goal after substitution and simplification.

Question 22. 3^{2n + 2} - 8n - 9, संख्या 8 से भाज्य है।
Answer: हल : मान लीजिए P(n): 3^{2n + 2} - 8n - 9 संख्या 8 से विभक्त होती है।
n = 1 के लिए,
3^{2 \(\cdot\) 1 + 2} - 8 \(\cdot\) 1 - 9 = 3⁴ - 8 - 9
= 81 - 17 = 64
जो 8 से विभाजित है।
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है अर्थात
3^{2k + 2} - 8k - 9, संख्या 8 से विभक्त होती है।
या 3^{2k + 2} - 8k - 9 = 8m
\(\therefore\) 3^{2k + 2} = 8m + 8k + 9
k को k + 1 से बदलने पर,
3^{2(k + 1) + 2} - 8(k + 1) - 9
= 3^{2k + 2 + 2} - 8k - 8 - 9
= 3² \(\cdot\) 3^{2k + 2} - 8k - 17
= 9(3^{2k + 2}) - 8k - 17
अब, 3^2k+2 = 8m + 8k + 9 रखने पर
= 9(8m + 8k + 9) - 8k - 17
= 72m + 72k + 81 - 8k - 17
= 72m + 64k + 64
= 8(9m + 8k + 8)
यह भी 8 से विभक्त होता है।
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves that \(3^{2n+2} - 8n - 9\) is divisible by 8 using mathematical induction. We verify the base case for n=1, which gives 64 (divisible by 8). Assuming it holds for n=k, we write \(3^{2k+2} - 8k - 9 = 8m\). Then, for n=k+1, we expand the expression, substitute \(3^{2k+2}\) using the P(k) assumption, and algebraically simplify to show that the entire expression is a multiple of 8.

🎯 Exam Tip: Divisibility proofs often require careful manipulation of exponents and strategic substitution of the inductive hypothesis. Ensure all remaining terms after substitution are clearly multiples of the divisor (8 in this case) to complete the proof.

Question 23. 41ⁿ - 14ⁿ, संख्या 27 का एक गुणज है।
Answer: हल : मान लीजिए P(n): 41ⁿ - 14ⁿ, संख्या 27 का गुणज है।
n = 1 के लिए, 41¹ - 14¹ = 41 - 14 = 27
\(\implies\) P(n), n = 1 के लिए सत्य है।
माना, P(n), n = k के लिए सत्य है।
\(\implies\) 41^k - 14^k = 27m
या 41^k = 27m + 14^k
k के स्थान पर k + 1 रखने पर
41^k+1 - 14^k+1
= 41 \(\cdot\) 41^k - 14^k+1
[41^k = 27m + 14^k रखने से]
= 41[27m + 14^k] - 14^k+1
= 41 \(\cdot\) 27m + 41 \(\cdot\) 14^k - 14 \(\cdot\) 14^k
= 41 \(\cdot\) 27m + 14^k (41 - 14)
= 41 \(\cdot\) 27m + 14^k \(\cdot\) 27
= 27[41m + 14^k]
जो कि 27 से विभक्त होता है।
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem proves that \(41^n - 14^n\) is divisible by 27 using mathematical induction. We verify the base case for n=1, which is 27 (divisible by 27). Assuming it holds for n=k, we write \(41^k = 27m + 14^k\). Then, for n=k+1, we expand \(41^{k+1} - 14^{k+1}\), substitute \(41^k\) from the P(k) assumption, and factor out 27 to demonstrate divisibility.

🎯 Exam Tip: The crucial step in this type of problem is to multiply the base of the higher power (41) with the inductive hypothesis and then factor out the common base (14) to reveal the divisor (27).

Question 24. (2n + 7) < (n + 3)².
Answer: हल : मान लीजिए P(n) = (2n + 7) < (n + 3)²
n = 1 के लिए बायाँ पक्ष = 2 \(\cdot\) 1 + 7 = 9
दायाँ पक्ष = (n + 3)² = (1 + 3)² = 4² = 16
9 < 16
\(\implies\) P(n), n = 1 के लिए सत्य है।
मान लीजिए P(n), n = k के लिए सत्य है।
\(\therefore\) 2k + 7 < (k + 3)²
[दोनों पक्षों में 2 जोड़ने से]
\(\implies\) 2k + 7 + 2 < (k + 3)² + 2
2k + 9 < k² + 6k + 9 + 2
2k + 9 < k² + 6k + 11 ...(1)
k को k + 1 रखने पर सिद्ध करना है।
2(k + 1) + 7 < (k + 1 + 3)²
2k + 2 + 7 < (k + 4)²
या 2k + 9 < (k + 4)²
समी. (1) में दाएँ पक्ष में 2k + 5 जोड़ने पर
2k + 9 < k² + 6k + 11
We need to show 2(k+1) + 7 < (k+1+3)²
i.e., 2k + 9 < (k+4)² = k² + 8k + 16
From P(k): 2k + 7 < (k+3)²
Add 2 to both sides: 2k + 9 < (k+3)² + 2
So, we need to show (k+3)² + 2 \(\le\) (k+4)²
(k+3)² + 2 = k² + 6k + 9 + 2 = k² + 6k + 11
(k+4)² = k² + 8k + 16
We need to prove: k² + 6k + 11 < k² + 8k + 16
Subtract k² from both sides: 6k + 11 < 8k + 16
Subtract 6k from both sides: 11 < 2k + 16
Subtract 16 from both sides: -5 < 2k
Since k \(\in\) N, k \(\ge\) 1, so 2k \(\ge\) 2. Thus, -5 < 2k is always true for k \(\in\) N.
Therefore, (k+3)² + 2 < (k+4)² for k \(\in\) N.
Since 2k + 9 < (k+3)² + 2 and (k+3)² + 2 < (k+4)², by transitivity, 2k + 9 < (k+4)². (This matches 2(k+1)+7 < (k+1+3)²)
\(\implies\) P(n), n = k + 1 के लिए भी सत्य है।
अतः गणितीय आगमन सिद्धांत के अनुसार P(n), n \(\in\) N, n के सभी मानों के लिए सत्य है।
In simple words: This problem uses mathematical induction to prove the inequality \(2n+7 < (n+3)^2\). We first verify the base case for n=1 (9 < 16, which is true). Assuming the inequality holds for n=k, we need to show that \(2(k+1)+7 < (k+1+3)^2\). By adding 2 to the P(k) inequality and comparing it with \((k+4)^2\), we demonstrate that \(2k+9\) is indeed less than \((k+4)^2\) for all natural numbers k, thus completing the inductive step.

🎯 Exam Tip: For inequality proofs, careful algebraic manipulation is crucial to show that the P(k+1) expression is indeed less than the required bound. It often involves showing that the increment on the left side is less than the increment on the right side, or directly comparing the simplified expressions.