CBSE Class 11 Mathematics Principle of Mathematical Induction Assignment Set 02

PRINCIPLE OF MATHEMATICAL INDUCTION

KEY POINTS

Induction and deduction are two basic processes of reasoning.

Deduction is the application of a general case to a particular case. In contrast to deduction, induction is process of reasoning from particular to general.

Principle of Mathematical Induction:
Let \(P(n)\) be any statement involving natural number \(n\) such that
(i) \(P(1)\) is true, and
(ii) If \(P(k)\) is true implies that \(P(k + 1)\) is also true for some natural number \(k\)
then \(P(n)\) is true for all \(n \in \mathbb{N}\)

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Using the principle of mathematical induction prove the following for all \(n \in \mathbb{N}\):

Question 1. \(3 \cdot 6 + 6 \cdot 9 + 9 \cdot 12 + \dots + 3n(3n + 3) = 3n(n + 1)(n + 2)\)
Answer:
Let the given statement be \(P(n)\):
\(P(n) : 3 \cdot 6 + 6 \cdot 9 + 9 \cdot 12 + \dots + 3n(3n + 3) = 3n(n + 1)(n + 2)\)
Step 1: For \(n = 1\),
LHS = \(3(1)(3(1) + 3) = 3(6) = 18\)
RHS = \(3(1)(1 + 1)(1 + 2) = 3(2)(3) = 18\)
Since LHS = RHS, \(P(1)\) is true.
Step 2: Assume that \(P(k)\) is true for some natural number \(k\), i.e.,
\(3 \cdot 6 + 6 \cdot 9 + 9 \cdot 12 + \dots + 3k(3k + 3) = 3k(k + 1)(k + 2)\)
Step 3: We need to prove that \(P(k + 1)\) is true, i.e.,
\(3 \cdot 6 + 6 \cdot 9 + \dots + 3k(3k + 3) + 3(k + 1)(3(k + 1) + 3) = 3(k + 1)(k + 2)(k + 3)\)
Consider LHS of \(P(k + 1)\):
\(= [3 \cdot 6 + 6 \cdot 9 + \dots + 3k(3k + 3)] + 3(k + 1)(3k + 3 + 3)\)
\(= 3k(k + 1)(k + 2) + 3(k + 1)(3k + 6)\)
\(= 3(k + 1) [k(k + 2) + (3k + 6)]\)
\(= 3(k + 1) [k^2 + 2k + 3k + 6]\)
\(= 3(k + 1) [k^2 + 5k + 6]\)
\(= 3(k + 1)(k + 2)(k + 3)\)
= RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by the Principle of Mathematical Induction, \(P(n)\) is true for all \(n \in \mathbb{N}\).
In simple words: We checked the statement for \(n=1\), assumed it holds for \(n=k\), and proved it holds for \(n=k+1\).

Exam Tip: Clearly substitute \(k+1\) for \(n\) in the last term and factor out common terms like \(3(k+1)\) to simplify algebraic manipulation.

Question 2. \(\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right)\dots\left(1 - \frac{1}{n + 1}\right) = \frac{1}{n + 1}\)
Answer:
Let \(P(n) : \left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\dots\left(1 - \frac{1}{n + 1}\right) = \frac{1}{n + 1}\)
Step 1: For \(n = 1\),
LHS = \(1 - \frac{1}{2} = \frac{1}{2}\)
RHS = \(\frac{1}{1 + 1} = \frac{1}{2}\)
LHS = RHS, so \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for \(n = k\), i.e.,
\(\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\dots\left(1 - \frac{1}{k + 1}\right) = \frac{1}{k + 1}\)
Step 3: We need to prove for \(n = k + 1\):
\(\left(1 - \frac{1}{2}\right)\dots\left(1 - \frac{1}{k + 1}\right)\left(1 - \frac{1}{(k + 1) + 1}\right) = \frac{1}{(k + 1) + 1}\)
Consider LHS:
\(= \left[\left(1 - \frac{1}{2}\right)\dots\left(1 - \frac{1}{k + 1}\right)\right] \cdot \left(1 - \frac{1}{k + 2}\right)\)
\(= \left(\frac{1}{k + 1}\right) \cdot \left(\frac{k + 2 - 1}{k + 2}\right)\)
\(= \left(\frac{1}{k + 1}\right) \cdot \left(\frac{k + 1}{k + 2}\right)\)
\(= \frac{1}{k + 2} = \frac{1}{(k + 1) + 1}\)
= RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Substitute the induction hypothesis into the product and cancel terms to get the required result.

Exam Tip: Telescopic products simplify easily when converted into fractions before multiplying.

Question 3. \(n^2 + n\) is an even natural number.
Answer:
Let \(P(n) : n^2 + n\) is an even natural number (i.e., \(n^2 + n = 2m\) for some integer \(m\)).
Step 1: For \(n = 1\),
\(1^2 + 1 = 1 + 1 = 2\), which is an even natural number.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(k^2 + k = 2m\) for some integer \(m\).
Step 3: We need to prove that \(P(k + 1)\) is true, i.e.,
\((k + 1)^2 + (k + 1)\) is even.
Consider:
\((k + 1)^2 + (k + 1) = k^2 + 2k + 1 + k + 1\)
\(= (k^2 + k) + 2k + 2\)
\(= 2m + 2(k + 1)\) (using induction hypothesis)
\(= 2(m + k + 1)\)
Since \(m + k + 1\) is an integer, \(2(m + k + 1)\) is an even number.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(n^2 + n\) is an even natural number for all \(n \in \mathbb{N}\).
In simple words: Group the terms to isolate \(k^2 + k\), substitute \(2m\), and factor out 2 to show divisibility.

Exam Tip: Expressing an even number as \(2 \times (\text{integer})\) is the standard way to prove divisibility by 2.

Question 4. \(2^{3n} - 1\) is divisible by 7
Answer:
Let \(P(n) : 2^{3n} - 1\) is divisible by 7 (i.e., \(2^{3n} - 1 = 7m\) for some integer \(m\)).
Step 1: For \(n = 1\),
\(2^{3(1)} - 1 = 2^3 - 1 = 8 - 1 = 7\), which is divisible by 7.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(2^{3k} - 1 = 7m\) for some integer \(m\), which gives \(2^{3k} = 7m + 1\).
Step 3: We need to prove that \(P(k + 1)\) is true, i.e.,
\(2^{3(k + 1)} - 1\) is divisible by 7.
Consider:
\(2^{3(k + 1)} - 1 = 2^{3k + 3} - 1 = 2^{3k} \cdot 2^3 - 1 = 8 \cdot 2^{3k} - 1\)
Substitute \(2^{3k} = 7m + 1\):
\(= 8(7m + 1) - 1\)
\(= 56m + 8 - 1\)
\(= 56m + 7 = 7(8m + 1)\)
Since \(8m + 1\) is an integer, \(7(8m + 1)\) is divisible by 7.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(2^{3n} - 1\) is divisible by 7 for all \(n \in \mathbb{N}\).
In simple words: Write \(2^{3(k+1)}\) as \(8 \cdot 2^{3k}\), substitute the assumed value, and show a factor of 7 remains.

Exam Tip: Isolate the exponential term from the assumption step to substitute cleanly into the next step.

Question 5. \(3^{2n}\) when divided by 8 leaves the remainder 1.
Answer:
Let \(P(n) : 3^{2n} = 8m + 1\) for some integer \(m \geq 0\).
Step 1: For \(n = 1\),
\(3^{2(1)} = 3^2 = 9 = 8(1) + 1\), which leaves a remainder of 1 when divided by 8.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(3^{2k} = 8m + 1\) for some integer \(m\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(3^{2(k + 1)} = 8p + 1\) for some integer \(p\).
Consider:
\(3^{2(k + 1)} = 3^{2k + 2} = 3^{2k} \cdot 3^2 = 9 \cdot 3^{2k}\)
Substitute \(3^{2k} = 8m + 1\):
\(= 9(8m + 1)\)
\(= 72m + 9\)
\(= 72m + 8 + 1\)
\(= 8(9m + 1) + 1\)
Let \(p = 9m + 1\), which is an integer. Thus, \(3^{2(k + 1)} = 8p + 1\).
This shows that \(3^{2(k + 1)}\) leaves a remainder of 1 when divided by 8.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(3^{2n}\) leaves a remainder 1 when divided by 8 for all \(n \in \mathbb{N}\).
In simple words: Express remainder statements in the form \(8m + 1\), then substitute and adjust constants to form a multiple of 8 plus 1.

Exam Tip: Splitting constants like \(9 = 8 + 1\) helps explicitly separate the remainder term \(+1\).

Question 6. \(4^n + 15n - 1\) is divisible by 9
Answer:
Let \(P(n) : 4^n + 15n - 1 = 9m\) for some integer \(m\).
Step 1: For \(n = 1\),
\(4^1 + 15(1) - 1 = 4 + 15 - 1 = 18\), which is divisible by 9.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(4^k + 15k - 1 = 9m\), which gives \(4^k = 9m - 15k + 1\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(4^{k + 1} + 15(k + 1) - 1\) is divisible by 9.
Consider:
\(4^{k + 1} + 15k + 15 - 1 = 4^k \cdot 4 + 15k + 14\)
Substitute \(4^k = 9m - 15k + 1\):
\(= 4(9m - 15k + 1) + 15k + 14\)
\(= 36m - 60k + 4 + 15k + 14\)
\(= 36m - 45k + 18\)
\(= 9(4m - 5k + 2)\)
Since \(4m - 5k + 2\) is an integer, the expression is divisible by 9.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(4^n + 15n - 1\) is divisible by 9 for all \(n \in \mathbb{N}\).
In simple words: Substitute \(4^k\) from the inductive hypothesis and simplify to reveal 9 as a common factor.

Exam Tip: Be careful with signs when distributing coefficients across multi-term algebraic substitutions.

Question 7. \(n^3 + (n + 1)^3 + (n + 2)^3\) is a multiple of 9.
Answer:
Let \(P(n) : n^3 + (n + 1)^3 + (n + 2)^3 = 9m\) for some integer \(m\).
Step 1: For \(n = 1\),
\(1^3 + (1 + 1)^3 + (1 + 2)^3 = 1 + 8 + 27 = 36\), which is a multiple of 9 (\(9 \times 4\)).
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(k^3 + (k + 1)^3 + (k + 2)^3 = 9m\) for some integer \(m\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\((k + 1)^3 + (k + 2)^3 + (k + 3)^3\) is a multiple of 9.
Consider:
\((k + 1)^3 + (k + 2)^3 + (k + 3)^3 = [(k + 1)^3 + (k + 2)^3 + k^3] - k^3 + (k + 3)^3\)
Using the assumption, \(k^3 + (k + 1)^3 + (k + 2)^3 = 9m\), so:
\(= 9m - k^3 + (k + 3)^3\)
Expand \((k + 3)^3 = k^3 + 9k^2 + 27k + 27\):
\(= 9m - k^3 + k^3 + 9k^2 + 27k + 27\)
\(= 9m + 9k^2 + 27k + 27\)
\(= 9(m + k^2 + 3k + 3)\)
Since \(m + k^2 + 3k + 3\) is an integer, the expression is a multiple of 9.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Add and subtract \(k^3\) to incorporate the induction hypothesis, then expand and factor out 9.

Exam Tip: Recognizing algebraic regrouping tricks saves time compared to expanding every cube from scratch.

Question 8. \(x^{2n - 1} - 1\) is divisible by \(x - 1, x \neq 1\)
Answer:
Let \(P(n) : x^{2n - 1} - 1\) is divisible by \(x - 1\).
Step 1: For \(n = 1\),
\(x^{2(1) - 1} - 1 = x^1 - 1 = x - 1\), which is clearly divisible by \(x - 1\).
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(x^{2k - 1} - 1 = (x - 1)m\) for some polynomial/integer \(m\), which gives \(x^{2k - 1} = (x - 1)m + 1\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(x^{2(k + 1) - 1} - 1 = x^{2k + 1} - 1\) is divisible by \(x - 1\).
Consider:
\(x^{2k + 1} - 1 = x^{2k - 1 + 2} - 1 = x^{2k - 1} \cdot x^2 - 1\)
Substitute \(x^{2k - 1} = (x - 1)m + 1\):
\(= [(x - 1)m + 1]x^2 - 1\)
\(= (x - 1)m x^2 + x^2 - 1\)
Factor \(x^2 - 1 = (x - 1)(x + 1)\):
\(= (x - 1)mx^2 + (x - 1)(x + 1)\)
\(= (x - 1)[mx^2 + x + 1]\)
Since the expression contains \(x - 1\) as a factor, it is divisible by \(x - 1\).
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(x^{2n - 1} - 1\) is divisible by \(x - 1\) for all \(n \in \mathbb{N}\).
In simple words: Express the exponent as \(2k-1 + 2\), substitute the assumption, and use the difference of squares formula.

Exam Tip: Factorize \(x^2 - 1\) immediately to extract the required divisor \(x - 1\).

Question 9. \(3^n > n\)
Answer:
Let \(P(n) : 3^n > n\).
Step 1: For \(n = 1\),
LHS = \(3^1 = 3\)
RHS = \(1\)
Since \(3 > 1\), \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(3^k > k\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(3^{k + 1} > k + 1\).
Consider LHS of \(P(k + 1)\):
\(3^{k + 1} = 3^k \cdot 3\)
Since \(3^k > k\), we have:
\(3^{k + 1} > 3k\)
For any natural number \(k \geq 1\), \(3k = k + 2k \geq k + 2 > k + 1\).
Thus, \(3^{k + 1} > k + 1\).
Therefore, \(P(k + 1)\) is true whenever \(P(k)\) is true.
By PMI, \(3^n > n\) for all \(n \in \mathbb{N}\).
In simple words: Multiply the inequality \(3^k > k\) by 3 and use inequality scaling to show it exceeds \(k+1\).

Exam Tip: For inequality proofs by induction, establish bounds by comparing scaled terms directly.

Question 10. If \(x\) and \(y\) are any two distinct integers then \(x^n - y^n\) is divisible by \((x - y)\)
Answer:
Let \(P(n) : x^n - y^n\) is divisible by \(x - y\) (i.e., \(x^n - y^n = (x - y)m\) for some integer \(m\)).
Step 1: For \(n = 1\),
\(x^1 - y^1 = x - y\), which is divisible by \(x - y\).
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(x^k - y^k = (x - y)m\), which gives \(x^k = (x - y)m + y^k\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(x^{k + 1} - y^{k + 1}\) is divisible by \(x - y\).
Consider:
\(x^{k + 1} - y^{k + 1} = x^k \cdot x - y^{k + 1}\)
Substitute \(x^k = (x - y)m + y^k\):
\(= [(x - y)m + y^k]x - y^{k + 1}\)
\(= (x - y)mx + y^k x - y^{k + 1}\)
\(= (x - y)mx + y^k(x - y)\)
\(= (x - y)[mx + y^k]\)
Since \(mx + y^k\) is an integer, the expression is divisible by \(x - y\).
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(x^n - y^n\) is divisible by \(x - y\) for all \(n \in \mathbb{N}\).
In simple words: Substitute \(x^k\) and factor out \((x - y)\) by grouping remaining terms with \(y^k\).

Exam Tip: This is a standard algebraic divisibility proof; master the substitution of \(x^k\) in terms of \(y^k\).

Question 11. \(n < 2^n\)
Answer:
Let \(P(n) : n < 2^n\).
Step 1: For \(n = 1\),
\(1 < 2^1\) i.e., \(1 < 2\), which is true.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(k < 2^k\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(k + 1 < 2^{k + 1}\).
Consider:
\(2^{k + 1} = 2^k \cdot 2 > k + k\) (since \(2^k > k\))\;
\(= 2k\)
For any natural number \(k \geq 1\), \(2k = k + k \geq k + 1\).
Thus, \(k + 1 < 2^{k + 1}\).
Therefore, \(P(k + 1)\) is true whenever \(P(k)\) is true.
By PMI, \(n < 2^n\) for all \(n \in \mathbb{N}\).
In simple words: Double the inequality \(2^k > k\) and show that \(2k\) is greater than or equal to \(k+1\).

Question 12. \(a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d]\)
Answer:
Let \(P(n) : a + (a + d) + (a + 2d) + \dots + [a + (n - 1)d] = \frac{n}{2}[2a + (n - 1)d]\)
Step 1: For \(n = 1\),
LHS = \(a\)
RHS = \(\frac{1}{2}[2a + (1 - 1)d] = \frac{1}{2}(2a) = a\)
LHS = RHS, so \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(a + (a + d) + \dots + [a + (k - 1)d] = \frac{k}{2}[2a + (k - 1)d]\)
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(a + (a + d) + \dots + [a + (k - 1)d] + [a + k d] = \frac{k + 1}{2}[2a + k d]\)
Consider LHS:
\(= \frac{k}{2}[2a + (k - 1)d] + [a + kd]\)
\(= \frac{k[2a + (k - 1)d] + 2[a + kd]}{2}\)
\(= \frac{2ak + k(k - 1)d + 2a + 2kd}{2}\)
\(= \frac{2a(k + 1) + d[k^2 - k + 2k]}{2}\)
\(= \frac{2a(k + 1) + d[k^2 + k]}{2}\)
\(= \frac{2a(k + 1) + dk(k + 1)}{2}\)
\(= \frac{k + 1}{2}[2a + kd]\)
= RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, the formula holds for all \(n \in \mathbb{N}\).
In simple words: Add the \((k+1)\)-th term of the arithmetic progression to the assumed sum and simplify algebraically.

Exam Tip: This is the standard summation formula for Arithmetic Progressions (AP) proven via induction.

Question 13. \(3x + 6x + 9x + \dots \text{ to } n \text{ terms} = \frac{3}{2}n(n + 1)x\)
Answer:
Let \(P(n) : 3x + 6x + 9x + \dots + 3nx = \frac{3}{2}n(n + 1)x\)
Step 1: For \(n = 1\),
LHS = \(3x\)
RHS = \(\frac{3}{2}(1)(1 + 1)x = \frac{3}{2}(2)x = 3x\)
LHS = RHS, so \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(3x + 6x + \dots + 3kx = \frac{3}{2}k(k + 1)x\)
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(3x + 6x + \dots + 3kx + 3(k + 1)x = \frac{3}{2}(k + 1)(k + 2)x\)
Consider LHS:
\(= \frac{3}{2}k(k + 1)x + 3(k + 1)x\)
\(= 3(k + 1)x \left[\frac{k}{2} + 1\right]\)
\(= 3(k + 1)x \left(\frac{k + 2}{2}\right)\)
\(= \frac{3}{2}(k + 1)(k + 2)x\)
= RHS
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, the statement is true for all \(n \in \mathbb{N}\).
In simple words: Factor out common terms like \(3(k+1)x\) from the sum to easily combine fractions.

Exam Tip: Factoring before finding common denominators simplifies series summation proofs significantly.

Question 14. \(11^{n + 2} + 12^{2n + 1}\) is divisible by 133.
Answer:
Let \(P(n) : 11^{n + 2} + 12^{2n + 1} = 133m\) for some integer \(m\).
Step 1: For \(n = 1\),
\(11^{1 + 2} + 12^{2(1) + 1} = 11^3 + 12^3 = 1331 + 1728 = 3059\)
Since \(3059 \div 133 = 23\), it is divisible by 133.
So, \(P(1)\) is true.
Step 2: Assume \(P(k)\) is true for some natural number \(k\), i.e.,
\(11^{k + 2} + 12^{2k + 1} = 133m\), which gives \(11^{k + 2} = 133m - 12^{2k + 1}\).
Step 3: We need to prove for \(n = k + 1\), i.e.,
\(11^{(k + 1) + 2} + 12^{2(k + 1) + 1} = 11^{k + 3} + 12^{2k + 3}\) is divisible by 133.
Consider:
\(11^{k + 3} + 12^{2k + 3} = 11^{k + 2} \cdot 11 + 12^{2k + 1} \cdot 12^2\)
Substitute \(11^{k + 2} = 133m - 12^{2k + 1}\):
\(= [133m - 12^{2k + 1}] \cdot 11 + 12^{2k + 1} \cdot 144\)
\(= 133m \cdot 11 - 11 \cdot 12^{2k + 1} + 144 \cdot 12^{2k + 1}\)
\(= 133(11m) + (144 - 11)12^{2k + 1}\)
\(= 133(11m) + 133 \cdot 12^{2k + 1}\)
\(= 133(11m + 12^{2k + 1})\)
Since \(11m + 12^{2k + 1}\) is an integer, the expression is divisible by 133.
Thus, \(P(k + 1)\) is true whenever \(P(k)\) is true.
Therefore, by PMI, \(11^{n + 2} + 12^{2n + 1}\) is divisible by 133 for all \(n \in \mathbb{N}\).
In simple words: Express \(11^{k+2}\) in terms of the assumption, substitute, and combine coefficients of \(12^{2k+1}\) to form 133.

Exam Tip: Notice how \(144 - 11 = 133\), which neatly completes the divisibility factor requirement.