OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Exercise 14 (I)

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(i)

Question 1. Find the sum to n terms of the series whose nth term is
(i) n(n + 2)
(ii) 3n² + 2n
(iii) 4n³ + 6n² + 2n
Answer:
(i) Given the nth term \( T_n = n(n + 2) \).
First, expand the expression: \( T_n = n^2 + 2n \).
To find the sum of n terms, \( S_n \), we need to sum \( T_n \).
\( S_n = \sum T_n = \sum (n^2 + 2n) = \sum n^2 + \sum 2n \).
Using standard formulas for sums of powers:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
So, \( S_n = \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} \).
Simplify the expression:
\( S_n = \frac{n(n+1)(2n+1)}{6} + n(n+1) \).
Take out the common factor \( \frac{n(n+1)}{6} \):
\( S_n = \frac{n(n+1)}{6} [ (2n+1) + 6 ] \).
\( S_n = \frac{n(n+1)}{6} (2n+7) \). This formula sums the terms efficiently.
(ii) Given the nth term \( T_n = 3n^2 + 2n \).
To find the sum of n terms, \( S_n \), we sum \( T_n \):
\( S_n = \sum T_n = \sum (3n^2 + 2n) = 3 \sum n^2 + 2 \sum n \).
Substitute the standard sum formulas:
\( S_n = 3 \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} \).
Simplify the expression:
\( S_n = \frac{n(n+1)(2n+1)}{2} + n(n+1) \).
Take out the common factor \( \frac{n(n+1)}{2} \):
\( S_n = \frac{n(n+1)}{2} [ (2n+1) + 2 ] \).
\( S_n = \frac{n(n+1)(2n+3)}{2} \). This is the sum of the series.
(iii) Given the nth term \( T_n = 4n^3 + 6n^2 + 2n \).
To find the sum of n terms, \( S_n \), we sum \( T_n \):
\( S_n = \sum T_n = \sum (4n^3 + 6n^2 + 2n) = 4 \sum n^3 + 6 \sum n^2 + 2 \sum n \).
Substitute the standard sum formulas:
\( \sum n^3 = \left[ \frac{n(n+1)}{2} \right]^2 = \frac{n^2(n+1)^2}{4} \)
\( S_n = 4 \frac{n^2(n+1)^2}{4} + 6 \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} \).
Simplify the expression:
\( S_n = n^2(n+1)^2 + n(n+1)(2n+1) + n(n+1) \).
Take out the common factor \( n(n+1) \):
\( S_n = n(n+1) [ n(n+1) + (2n+1) + 1 ] \).
Simplify the terms inside the square bracket:
\( S_n = n(n+1) [ n^2 + n + 2n + 1 + 1 ] \).
\( S_n = n(n+1) [ n^2 + 3n + 2 ] \).
Factor the quadratic expression \( n^2 + 3n + 2 \):
\( n^2 + 3n + 2 = (n+1)(n+2) \).
So, \( S_n = n(n+1) [ (n+1)(n+2) ] \).
\( S_n = n(n+1)^2(n+2) \). This is the sum of the series.
In simple words: For each part, we take the given formula for the nth term, expand it, and then use the known formulas for the sum of n, n², and n³ terms. Then, we simplify the whole expression by finding common factors.

🎯 Exam Tip: Remember the standard sum formulas for \( \sum n \), \( \sum n^2 \), and \( \sum n^3 \). Factoring out common terms early in the simplification process can save a lot of time and reduce calculation errors.

Question 2. Find the sum of the series
(i) 3 × 5 + 5 × 7 + 7 × 9 + ..... to n terms
(ii) 1² + 3² + 5² +..... to n terms
(iii) 2² + 4² + 6² + ..... to n terms.
Answer:
(i) The series is \( 3 \times 5 + 5 \times 7 + 7 \times 9 + \dots \) to n terms.
First, find the nth term of the sequence \( 3, 5, 7, \dots \). This is an arithmetic progression (AP) with first term \( a=3 \) and common difference \( d=2 \).
The nth term is \( a + (n-1)d = 3 + (n-1)2 = 3 + 2n - 2 = 2n + 1 \).
Next, find the nth term of the sequence \( 5, 7, 9, \dots \). This is an AP with first term \( a=5 \) and common difference \( d=2 \).
The nth term is \( a + (n-1)d = 5 + (n-1)2 = 5 + 2n - 2 = 2n + 3 \).
So, the nth term of the series \( T_n \) is the product of these two terms:
\( T_n = (2n + 1)(2n + 3) \).
Expand \( T_n \): \( T_n = 4n^2 + 6n + 2n + 3 = 4n^2 + 8n + 3 \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (4n^2 + 8n + 3) = 4 \sum n^2 + 8 \sum n + \sum 3 \).
Substitute the standard sum formulas:
\( S_n = 4 \frac{n(n+1)(2n+1)}{6} + 8 \frac{n(n+1)}{2} + 3n \).
Simplify the expression:
\( S_n = \frac{2n(n+1)(2n+1)}{3} + 4n(n+1) + 3n \).
Take out the common factor \( \frac{n}{3} \):
\( S_n = \frac{n}{3} [ 2(n+1)(2n+1) + 12(n+1) + 9 ] \).
Simplify the terms inside the square bracket:
\( S_n = \frac{n}{3} [ 2(2n^2 + 3n + 1) + 12n + 12 + 9 ] \).
\( S_n = \frac{n}{3} [ 4n^2 + 6n + 2 + 12n + 21 ] \).
\( S_n = \frac{n}{3} [ 4n^2 + 18n + 23 ] \). This is the sum of the series.
(ii) The series is \( 1^2 + 3^2 + 5^2 + \dots \) to n terms.
The numbers \( 1, 3, 5, \dots \) form an AP with \( a=1 \) and \( d=2 \).
The nth term of this AP is \( a + (n-1)d = 1 + (n-1)2 = 1 + 2n - 2 = 2n - 1 \).
So, the nth term of the series \( T_n = (2n - 1)^2 \).
Expand \( T_n \): \( T_n = (2n)^2 - 2(2n)(1) + 1^2 = 4n^2 - 4n + 1 \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (4n^2 - 4n + 1) = 4 \sum n^2 - 4 \sum n + \sum 1 \).
Substitute the standard sum formulas:
\( S_n = 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n \).
Simplify the expression:
\( S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \).
Take out the common factor \( \frac{n}{3} \):
\( S_n = \frac{n}{3} [ 2(n+1)(2n+1) - 6(n+1) + 3 ] \).
Simplify the terms inside the square bracket:
\( S_n = \frac{n}{3} [ 2(2n^2 + 3n + 1) - 6n - 6 + 3 ] \).
\( S_n = \frac{n}{3} [ 4n^2 + 6n + 2 - 6n - 3 ] \).
\( S_n = \frac{n}{3} [ 4n^2 - 1 ] \).
This can be further factored using \( a^2 - b^2 = (a-b)(a+b) \):
\( S_n = \frac{n}{3} (2n - 1)(2n + 1) \). This is the sum of the series.
(iii) The series is \( 2^2 + 4^2 + 6^2 + \dots \) to n terms.
The numbers \( 2, 4, 6, \dots \) form an AP with \( a=2 \) and \( d=2 \).
The nth term of this AP is \( a + (n-1)d = 2 + (n-1)2 = 2 + 2n - 2 = 2n \).
So, the nth term of the series \( T_n = (2n)^2 = 4n^2 \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (4n^2) = 4 \sum n^2 \).
Substitute the standard sum formula:
\( S_n = 4 \frac{n(n+1)(2n+1)}{6} \).
Simplify the expression:
\( S_n = \frac{2n(n+1)(2n+1)}{3} \). This is the sum of the series.
In simple words: First, we find a general formula for the nth term of the series. Then we expand this formula. After that, we use the known formulas for summing powers of n to find the total sum up to n terms. Finally, we simplify the result.

🎯 Exam Tip: Always identify the pattern of the terms first to find the correct nth term formula. Be careful with algebraic expansions and factorizations, as small errors can lead to incorrect final sums.

Question 3. Find the nth term and the sum to n terms of the series 1.2 + 2.3 + 3.4 + .....
Answer:
The given series is \( 1 \times 2 + 2 \times 3 + 3 \times 4 + \dots \).
First, find the nth term of the first sequence \( 1, 2, 3, \dots \). The nth term is simply \( n \).
Next, find the nth term of the second sequence \( 2, 3, 4, \dots \). The nth term is \( n+1 \).
So, the nth term of the series \( T_n = n(n+1) \).
Expand \( T_n \): \( T_n = n^2 + n \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (n^2 + n) = \sum n^2 + \sum n \).
Substitute the standard sum formulas:
\( S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \).
Take out the common factor \( \frac{n(n+1)}{6} \):
\( S_n = \frac{n(n+1)}{6} [ (2n+1) + 3 ] \).
\( S_n = \frac{n(n+1)}{6} (2n+4) \).
Factor out 2 from \( (2n+4) \):
\( S_n = \frac{n(n+1)}{6} [2(n+2)] \).
\( S_n = \frac{2n(n+1)(n+2)}{6} \).
\( S_n = \frac{n(n+1)(n+2)}{3} \). This is the sum of the series.
In simple words: We find the formula for the nth term by looking at the patterns in each part of the product. Then we sum this nth term using common formulas for sums of powers, and simplify the result.

🎯 Exam Tip: When terms are products, find the nth term of each factor separately, then multiply them to get the series' nth term. This method simplifies finding the overall sum.

Question 4. Sum up to n terms the series 1.2² + 2.3² + 3.4² + .....
Answer:
The given series is \( 1 \times 2^2 + 2 \times 3^2 + 3 \times 4^2 + \dots \).
First, find the nth term of the first sequence \( 1, 2, 3, \dots \). The nth term is \( n \).
Next, find the nth term of the sequence \( 2, 3, 4, \dots \). The nth term is \( n+1 \).
So, the nth term of the series \( T_n = n(n+1)^2 \).
Expand \( T_n \):
\( T_n = n(n^2 + 2n + 1) \).
\( T_n = n^3 + 2n^2 + n \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (n^3 + 2n^2 + n) = \sum n^3 + 2 \sum n^2 + \sum n \).
Substitute the standard sum formulas:
\( S_n = \frac{n^2(n+1)^2}{4} + 2 \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \).
Simplify the expression:
\( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{3} + \frac{n(n+1)}{2} \).
To add these fractions, find a common denominator, which is 12. Take out the common factor \( \frac{n(n+1)}{12} \):
\( S_n = \frac{n(n+1)}{12} [ 3n(n+1) + 4(2n+1) + 6 ] \).
Simplify the terms inside the square bracket:
\( S_n = \frac{n(n+1)}{12} [ 3n^2 + 3n + 8n + 4 + 6 ] \).
\( S_n = \frac{n(n+1)}{12} [ 3n^2 + 11n + 10 ] \).
Factor the quadratic expression \( 3n^2 + 11n + 10 \).
We look for two numbers that multiply to \( 3 \times 10 = 30 \) and add to 11. These are 5 and 6.
\( 3n^2 + 6n + 5n + 10 = 3n(n+2) + 5(n+2) = (n+2)(3n+5) \).
So, \( S_n = \frac{n(n+1)(n+2)(3n+5)}{12} \). This is the sum of the series.
In simple words: We first find the general formula for any term in the series by looking at its pattern. Then we expand this formula and use known mathematical rules for summing up powers of numbers. Finally, we simplify the result to get the total sum.

🎯 Exam Tip: Always expand \( (n+1)^2 \) correctly as \( n^2+2n+1 \) to avoid common algebraic errors. When adding fractions, finding the least common multiple for the denominators is key for simplification.

Question 5. Sum up 1 +(1 + 2) + (1 + 2 + 3) + ....... +(1 + 2 + 3 + ..... + n).
Answer:
The given series is \( 1 +(1 + 2) + (1 + 2 + 3) + \dots + (1 + 2 + 3 + \dots + n) \).
The nth term of this series, \( T_n \), is the sum of the first n natural numbers.
We know that the sum of the first \( k \) natural numbers is \( \frac{k(k+1)}{2} \).
So, for the nth term, \( T_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \).
We can also write this as \( T_n = \frac{1}{2} (n^2 + n) \).
Now, to find the sum of n terms of the given series, \( S_n = \sum T_n \):
\( S_n = \sum \frac{1}{2} (n^2 + n) = \frac{1}{2} (\sum n^2 + \sum n) \).
Substitute the standard sum formulas:
\( S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \).
Take out the common factor \( \frac{n(n+1)}{2} \) from inside the bracket:
\( S_n = \frac{1}{2} \left[ \frac{n(n+1)}{2} \left( \frac{2n+1}{3} + 1 \right) \right] \).
\( S_n = \frac{n(n+1)}{4} \left[ \frac{2n+1+3}{3} \right] \).
\( S_n = \frac{n(n+1)}{4} \left[ \frac{2n+4}{3} \right] \).
Factor out 2 from \( (2n+4) \):
\( S_n = \frac{n(n+1)}{4} \frac{2(n+2)}{3} \).
\( S_n = \frac{2n(n+1)(n+2)}{12} \).
\( S_n = \frac{n(n+1)(n+2)}{6} \). This is the sum of the series. The triangular numbers form the basis for this type of sum.
In simple words: Each term in the series is itself a sum of natural numbers. We first find a simple formula for each term. Then, we use the known rules to sum up these terms, which gives us the final answer.

🎯 Exam Tip: Recognize patterns quickly. The sum of the first 'n' natural numbers, often called a triangular number, is a common building block for more complex series. Be careful with fractional coefficients during simplification.

Question 6. Sum up to n terms the series 1 + \left(1+\frac{1}{2}\right) + \left(1+\frac{1}{2}+\frac{1}{4}\right) ....
Answer:
The given series is \( 1 + \left(1+\frac{1}{2}\right) + \left(1+\frac{1}{2}+\frac{1}{4}\right) + \dots \).
Let's find the nth term, \( T_n \). Each term is a sum of a geometric progression (GP).
The terms inside the bracket are of the form \( 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-1}} \).
This is a GP with first term \( a=1 \) and common ratio \( r = \frac{1}{2} \).
The sum of n terms of a GP is \( S_k = \frac{a(1-r^k)}{1-r} \). Here, \( k \) is the number of terms in the inner sum, which is \( n \).
So, \( T_n = \frac{1 \left(1 - \left(\frac{1}{2}\right)^n \right)}{1 - \frac{1}{2}} \).
\( T_n = \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} \).
\( T_n = 2 \left(1 - \frac{1}{2^n}\right) \).
\( T_n = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}} \).
Now, find the sum of the series to n terms, \( S_n = \sum T_n \):
\( S_n = \sum \left(2 - \frac{1}{2^{n-1}}\right) \).
\( S_n = \sum 2 - \sum \frac{1}{2^{n-1}} \).
\( S_n = 2n - \left( 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n-1}} \right) \).
The second part is a GP with \( a=1 \), \( r=\frac{1}{2} \), and \( n \) terms.
The sum of this GP is \( \frac{1(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} = 2 \left(1 - \frac{1}{2^n}\right) \).
So, \( S_n = 2n - 2 \left(1 - \frac{1}{2^n}\right) \).
\( S_n = 2n - 2 + \frac{2}{2^n} \).
\( S_n = 2n - 2 + \frac{1}{2^{n-1}} \). This is the sum of the series. This type of series involves both arithmetic and geometric progression concepts.
In simple words: Each part of the main series is a sum of a geometric progression. We first find the formula for this inner sum, which becomes the general term for our main series. Then we sum these general terms to find the total sum.

🎯 Exam Tip: When terms of a series are themselves sums (or products), first find the nth term of the inner sum/product. Recognize if it's an AP or GP. Carefully apply the sum formulas for AP/GP before summing the main series.

Question 7. Sum up to n terms the series where nth term is 2n - 1.
Answer:
The given nth term is \( T_n = 2^n - 1 \).
To find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (2^n - 1) = \sum 2^n - \sum 1 \).
The \( \sum 2^n \) part is the sum of a geometric progression: \( 2^1 + 2^2 + 2^3 + \dots + 2^n \).
In this GP, the first term \( a=2 \) and the common ratio \( r=2 \). Since \( r > 1 \), the sum of n terms is \( \frac{a(r^n - 1)}{r-1} \).
So, \( \sum 2^n = \frac{2(2^n - 1)}{2-1} = \frac{2(2^n - 1)}{1} = 2(2^n - 1) \).
The \( \sum 1 \) part is simply \( n \).
Substitute these back into the expression for \( S_n \):
\( S_n = 2(2^n - 1) - n \).
\( S_n = 2^{n+1} - 2 - n \). This is the sum of the series. Recognizing the geometric part is key here.
In simple words: We are given the formula for the nth term. We break it into two parts: a geometric series and a constant term. We sum each part separately using their specific sum formulas and then combine them to get the final answer.

🎯 Exam Tip: Clearly separate terms when summing. For \( \sum (A_n - B_n) \), it's \( \sum A_n - \sum B_n \). Recognize the geometric progression immediately and apply its sum formula correctly, paying attention to whether the common ratio is greater or less than 1.

Question 8. Sum up 1 + 4 + 8 + 13 + ..... to n terms.
Answer:
The given series is \( 1 + 4 + 8 + 13 + \dots \) to n terms.
Let \( S_n = 1 + 4 + 8 + 13 + \dots + T_{n-1} + T_n \) ...(1)
Write the series again, shifting terms one place to the right:
\( S_n = 0 + 1 + 4 + 8 + \dots + T_{n-1} \) ...(2)
Subtract (2) from (1):
\( S_n - S_n = (1 + 4 + 8 + 13 + \dots + T_{n-1} + T_n) - (0 + 1 + 4 + 8 + \dots + T_{n-1}) \).
\( 0 = 1 + (4-1) + (8-4) + (13-8) + \dots + (T_n - T_{n-1}) - T_n \).
\( 0 = 1 + 3 + 4 + 5 + \dots + (n-1) \text{ terms} - T_n \).
So, \( T_n = 1 + (3 + 4 + 5 + \dots \text{ up to } (n-1) \text{ terms}) \).
The terms in the parenthesis form an arithmetic progression (AP) with first term \( a=3 \), common difference \( d=1 \), and number of terms \( k = n-1 \).
The sum of an AP is \( \frac{k}{2} [2a + (k-1)d] \).
So, the sum of terms in parenthesis is \( \frac{n-1}{2} [2(3) + (n-1-1)(1)] \).
\( = \frac{n-1}{2} [6 + (n-2)] \).
\( = \frac{n-1}{2} (n+4) \).
Substitute this back into the expression for \( T_n \):
\( T_n = 1 + \frac{(n-1)(n+4)}{2} \).
\( T_n = \frac{2 + (n-1)(n+4)}{2} \).
\( T_n = \frac{2 + n^2 + 4n - n - 4}{2} \).
\( T_n = \frac{n^2 + 3n - 2}{2} \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum \frac{n^2 + 3n - 2}{2} = \frac{1}{2} (\sum n^2 + 3 \sum n - \sum 2) \).
Substitute the standard sum formulas:
\( S_n = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + 3 \frac{n(n+1)}{2} - 2n \right] \).
Take out the common factor \( \frac{n}{6} \) from inside the bracket (to make a common denominator of 6):
\( S_n = \frac{n}{12} [ (n+1)(2n+1) + 9(n+1) - 12 ] \).
Simplify the terms inside the square bracket:
\( S_n = \frac{n}{12} [ (2n^2 + 3n + 1) + (9n + 9) - 12 ] \).
\( S_n = \frac{n}{12} [ 2n^2 + 12n - 2 ] \).
Factor out 2 from the bracket:
\( S_n = \frac{n}{12} [ 2(n^2 + 6n - 1) ] \).
\( S_n = \frac{n}{6} (n^2 + 6n - 1) \). This is the sum of the series. This method involves using differences to find the general term.
In simple words: We find the pattern of differences between consecutive terms to work out the formula for the nth term. Once we have the nth term, we use the standard sum formulas to find the sum of the entire series.

🎯 Exam Tip: The method of differences is crucial for series that are not standard AP or GP. Write out \( S_n \) twice, shifted, and subtract to find \( T_n \). Be careful with the number of terms when summing the difference series.

Question 9. Sum up 3 + 5 + 11 + 29 + ..... to n terms.
Answer:
The given series is \( 3 + 5 + 11 + 29 + \dots \) to n terms.
Let \( S_n = 3 + 5 + 11 + 29 + \dots + T_{n-1} + T_n \) ...(1)
Write the series again, shifting terms one place to the right:
\( S_n = 0 + 3 + 5 + 11 + \dots + T_{n-1} \) ...(2)
Subtract (2) from (1):
\( 0 = 3 + (5-3) + (11-5) + (29-11) + \dots + (T_n - T_{n-1}) - T_n \).
\( 0 = 3 + 2 + 6 + 18 + \dots + (T_n - T_{n-1}) - T_n \).
So, \( T_n = 3 + (2 + 6 + 18 + \dots \text{ up to } (n-1) \text{ terms}) \).
The terms in the parenthesis form a geometric progression (GP) with first term \( a=2 \), common ratio \( r=\frac{6}{2}=3 \), and number of terms \( k = n-1 \).
The sum of a GP is \( \frac{a(r^k - 1)}{r-1} \). Since \( r > 1 \), we use this formula.
So, the sum of terms in parenthesis is \( \frac{2(3^{n-1} - 1)}{3-1} \).
\( = \frac{2(3^{n-1} - 1)}{2} = 3^{n-1} - 1 \).
Substitute this back into the expression for \( T_n \):
\( T_n = 3 + (3^{n-1} - 1) \).
\( T_n = 3^{n-1} + 2 \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (3^{n-1} + 2) = \sum 3^{n-1} + \sum 2 \).
The \( \sum 3^{n-1} \) part is the sum of a GP: \( 3^0 + 3^1 + 3^2 + \dots + 3^{n-1} \).
In this GP, the first term \( a=1 \) (since \( 3^0=1 \)), and the common ratio \( r=3 \). There are \( n \) terms (from \( n=1 \) to \( n \)).
So, \( \sum 3^{n-1} = \frac{1(3^n - 1)}{3-1} = \frac{3^n - 1}{2} \).
The \( \sum 2 \) part is simply \( 2n \).
Substitute these back into the expression for \( S_n \):
\( S_n = \frac{3^n - 1}{2} + 2n \). This is the sum of the series. Identifying the geometric progression in the differences is crucial.
In simple words: We first find the difference between each term to spot a pattern. This pattern turns out to be a geometric series. We use its sum formula to find the general nth term. Then, we sum these nth terms to get the total sum of the original series.

🎯 Exam Tip: Pay close attention to the starting term and number of terms in the difference series, especially if it's a geometric progression starting from \( r^0 \). The common ratio must be determined correctly. Check the first few terms to verify your nth term formula.

Question 10. Sum to n terms the series 7 + 77 + 777 + ....
Answer:
The given series is \( 7 + 77 + 777 + \dots \) to n terms.
Let \( S_n = 7 + 77 + 777 + \dots + T_n \).
First, factor out 7 from each term:
\( S_n = 7 (1 + 11 + 111 + \dots + \text{nth term}) \).
Next, multiply and divide by 9 (a common trick for this type of series):
\( S_n = \frac{7}{9} (9 + 99 + 999 + \dots + \text{nth term}) \).
Rewrite each term inside the parenthesis:
\( 9 = 10 - 1 \)
\( 99 = 100 - 1 = 10^2 - 1 \)
\( 999 = 1000 - 1 = 10^3 - 1 \)
So, the nth term inside the parenthesis is \( 10^n - 1 \).
Therefore, \( S_n = \frac{7}{9} \sum (10^n - 1) \).
\( S_n = \frac{7}{9} \left( \sum 10^n - \sum 1 \right) \).
The \( \sum 10^n \) part is the sum of a geometric progression: \( 10^1 + 10^2 + 10^3 + \dots + 10^n \).
In this GP, the first term \( a=10 \) and the common ratio \( r=10 \). There are \( n \) terms.
The sum of this GP is \( \frac{a(r^n - 1)}{r-1} = \frac{10(10^n - 1)}{10-1} = \frac{10(10^n - 1)}{9} \).
The \( \sum 1 \) part is simply \( n \).
Substitute these back into the expression for \( S_n \):
\( S_n = \frac{7}{9} \left( \frac{10(10^n - 1)}{9} - n \right) \).
\( S_n = \frac{70}{81} (10^n - 1) - \frac{7n}{9} \). This is the sum of the series. This method is effective for repetitive digit sequences.
In simple words: For series with repeating digits, we factor out the digit, then multiply and divide by 9. This changes the terms into expressions like \( 10^n - 1 \). We then sum these new terms using geometric series formulas and simplify.

🎯 Exam Tip: The trick of multiplying by 9 is essential for this type of series (e.g., 6+66+666 or 0.3+0.33+0.333). Always remember to factor out the repeated digit first, then handle the \( 10^n - 1 \) pattern carefully. Pay attention to the initial term in the geometric sum.

Question 11. Sum to n terms the series 1 + 3 + 7 + 15 + 31 + ....
Answer:
The given series is \( 1 + 3 + 7 + 15 + 31 + \dots \) to n terms.
Let \( S_n = 1 + 3 + 7 + 15 + \dots + T_{n-1} + T_n \) ...(1)
Write the series again, shifting terms one place to the right:
\( S_n = 0 + 1 + 3 + 7 + \dots + T_{n-1} \) ...(2)
Subtract (2) from (1):
\( 0 = 1 + (3-1) + (7-3) + (15-7) + (31-15) + \dots + (T_n - T_{n-1}) - T_n \).
\( 0 = 1 + 2 + 4 + 8 + 16 + \dots + (T_n - T_{n-1}) - T_n \).
So, \( T_n = 1 + (2 + 4 + 8 + 16 + \dots \text{ up to } (n-1) \text{ terms}) \).
The terms in the parenthesis form a geometric progression (GP) with first term \( a=2 \), common ratio \( r=2 \), and number of terms \( k = n-1 \).
The sum of this GP is \( \frac{a(r^k - 1)}{r-1} = \frac{2(2^{n-1} - 1)}{2-1} = 2(2^{n-1} - 1) \).
\( = 2^n - 2 \).
Substitute this back into the expression for \( T_n \):
\( T_n = 1 + (2^n - 2) \).
\( T_n = 2^n - 1 \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (2^n - 1) = \sum 2^n - \sum 1 \).
The \( \sum 2^n \) part is a GP: \( 2^1 + 2^2 + \dots + 2^n \). Here \( a=2, r=2 \).
Sum is \( \frac{2(2^n - 1)}{2-1} = 2(2^n - 1) = 2^{n+1} - 2 \).
The \( \sum 1 \) part is \( n \).
So, \( S_n = (2^{n+1} - 2) - n \).
\( S_n = 2^{n+1} - n - 2 \). This is the sum of the series. The method of differences helps in finding the general term.
In simple words: We use the method of differences to find the formula for the nth term of the series. It turns out the differences form a geometric series. Once we have the nth term, we sum it up by separating it into a geometric sum and a simple sum, then combine these results.

🎯 Exam Tip: The method of differences is effective when the series is not a direct AP or GP. Carefully identify the type of series formed by the differences, whether it's an AP, GP, or another known sequence, to find the nth term accurately.

Question 12. Find the sum to n terms of the series 1. 2. 3 + 2. 3. 4 + 3. 4. 5 + ....
Answer:
The given series is \( 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + \dots \).
First, find the nth term, \( T_n \). Each term is a product of three consecutive integers.
The first factor sequence is \( 1, 2, 3, \dots \), whose nth term is \( n \).
The second factor sequence is \( 2, 3, 4, \dots \), whose nth term is \( n+1 \).
The third factor sequence is \( 3, 4, 5, \dots \), whose nth term is \( n+2 \).
So, the nth term of the series \( T_n = n(n+1)(n+2) \).
Expand \( T_n \):
\( T_n = n(n^2 + 3n + 2) \).
\( T_n = n^3 + 3n^2 + 2n \).
Now, find the sum to n terms, \( S_n = \sum T_n \):
\( S_n = \sum (n^3 + 3n^2 + 2n) = \sum n^3 + 3 \sum n^2 + 2 \sum n \).
Substitute the standard sum formulas:
\( S_n = \frac{n^2(n+1)^2}{4} + 3 \frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2} \).
Simplify the expression:
\( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) \).
Take out the common factor \( \frac{n(n+1)}{4} \):
\( S_n = \frac{n(n+1)}{4} [ n(n+1) + 2(2n+1) + 4 ] \).
Simplify the terms inside the square bracket:
\( S_n = \frac{n(n+1)}{4} [ n^2 + n + 4n + 2 + 4 ] \).
\( S_n = \frac{n(n+1)}{4} [ n^2 + 5n + 6 ] \).
Factor the quadratic expression \( n^2 + 5n + 6 \):
\( n^2 + 5n + 6 = (n+2)(n+3) \).
So, \( S_n = \frac{n(n+1)(n+2)(n+3)}{4} \). This is the sum of the series. This type of sum of products is often simplified by the method of differences or by using known formulas for sums of powers.
In simple words: We find the general formula for each term in the series by observing the pattern of the multiplied numbers. We then expand this formula and use known rules for summing powers of n. Finally, we simplify the resulting expression to get the total sum.

🎯 Exam Tip: For products of consecutive numbers, the sum often follows a pattern where the product extends to one more term, divided by the number of terms. For example, \( \sum n(n+1)(n+2) = \frac{n(n+1)(n+2)(n+3)}{4} \).

Question 13. Find the sum of the series to n terms and to infinity: \frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + ....
Answer:
The given series is \( \frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + \dots \).
First, find the nth term, \( T_n \).
The first factors are \( 1, 3, 5, \dots \), which form an AP with \( a=1, d=2 \). Its nth term is \( 1 + (n-1)2 = 2n-1 \).
The second factors are \( 3, 5, 7, \dots \), which form an AP with \( a=3, d=2 \). Its nth term is \( 3 + (n-1)2 = 2n+1 \).
So, the nth term of the series is \( T_n = \frac{1}{(2n-1)(2n+1)} \).
This can be written using partial fractions. We aim to express \( T_n \) as a difference of two terms, \( A \frac{1}{2n-1} + B \frac{1}{2n+1} \).
Let \( \frac{1}{(2n-1)(2n+1)} = \frac{A}{2n-1} + \frac{B}{2n+1} \).
Multiply both sides by \( (2n-1)(2n+1) \):
\( 1 = A(2n+1) + B(2n-1) \).
If \( 2n-1=0 \implies n=\frac{1}{2} \): \( 1 = A(1+1) + B(0) \implies 1 = 2A \implies A = \frac{1}{2} \).
If \( 2n+1=0 \implies n=-\frac{1}{2} \): \( 1 = A(0) + B(-1-1) \implies 1 = -2B \implies B = -\frac{1}{2} \).
So, \( T_n = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \). This is a telescoping series.
Now, find the sum to n terms, \( S_n = \sum T_n \). Let's write out the first few terms:
\( T_1 = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} \right) \)
\( T_2 = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \)
\( T_3 = \frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right) \)
...
\( T_n = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \)
When we sum these terms, the middle terms cancel out (telescoping sum):
\( S_n = T_1 + T_2 + \dots + T_n = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \right] \).
\( S_n = \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) \).
\( S_n = \frac{1}{2} \left( \frac{2n+1 - 1}{2n+1} \right) \).
\( S_n = \frac{1}{2} \left( \frac{2n}{2n+1} \right) = \frac{n}{2n+1} \). This is the sum of the series to n terms.
Now, find the sum to infinity, \( S_\infty \). This is found by taking the limit of \( S_n \) as \( n \to \infty \):
\( S_\infty = \lim_{n \to \infty} \frac{n}{2n+1} \).
Divide numerator and denominator by \( n \):
\( S_\infty = \lim_{n \to \infty} \frac{1}{2 + \frac{1}{n}} \).
As \( n \to \infty \), \( \frac{1}{n} \to 0 \).
So, \( S_\infty = \frac{1}{2 + 0} = \frac{1}{2} \). The series converges to this value.
In simple words: First, we find the general formula for each term. Then, we use a technique called partial fractions to rewrite each term as a subtraction of two simpler parts. When we add these terms together, most of them cancel out, leaving a simple formula for the sum up to 'n' terms. To find the sum to infinity, we see what this sum approaches as 'n' gets very, very large.

🎯 Exam Tip: Series of the form \( \frac{1}{(an+b)(an+c)} \) are often solved using partial fractions to create a telescoping sum. Remember that \( \lim_{n \to \infty} \frac{1}{n} = 0 \) when calculating the sum to infinity.

Question 14. Sum to n terms the series whose nth term is \frac{1}{4 n^2-1}
Answer:
The given nth term is \( T_n = \frac{1}{4n^2 - 1} \).
First, factor the denominator using the difference of squares formula, \( a^2 - b^2 = (a-b)(a+b) \):
\( 4n^2 - 1 = (2n)^2 - 1^2 = (2n-1)(2n+1) \).
So, \( T_n = \frac{1}{(2n-1)(2n+1)} \).
This is the same nth term as in Question 13. We will use partial fractions to split this term.
\( T_n = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \).
Now, find the sum to n terms, \( S_n = \sum T_n \). We write out the first few terms to see the telescoping pattern:
\( T_1 = \frac{1}{2} \left( \frac{1}{1} - \frac{1}{3} \right) \)
\( T_2 = \frac{1}{2} \left( \frac{1}{3} - \frac{1}{5} \right) \)
\( T_3 = \frac{1}{2} \left( \frac{1}{5} - \frac{1}{7} \right) \)
...
\( T_n = \frac{1}{2} \left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \)
When we sum these terms, the middle terms cancel out:
\( S_n = \frac{1}{2} \left[ \left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \dots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \right] \).
\( S_n = \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) \).
\( S_n = \frac{1}{2} \left( \frac{2n+1 - 1}{2n+1} \right) \).
\( S_n = \frac{1}{2} \left( \frac{2n}{2n+1} \right) = \frac{n}{2n+1} \). This is the sum of the series to n terms. This solution shows how common partial fraction forms appear.
In simple words: We first rewrite the given nth term using a special algebra trick called factoring the bottom part. This allows us to split the term into a subtraction of two simpler fractions. When we add up these new terms, most of them cancel out, leaving us with a simple formula for the total sum.

🎯 Exam Tip: Always look for factoring opportunities in the denominator, especially for expressions like \( a^2 - b^2 \). This often simplifies the problem into a telescoping series, making summation straightforward.

Question 15. Natural numbers are written as
1
2 3
4 5 6
.......
Show that the sum of the numbers in the nth group is \frac{n}{2}(n² + 1).
Answer:
The groups of natural numbers are arranged as follows:
Group 1: 1 (1 term)
Group 2: 2, 3 (2 terms)
Group 3: 4, 5, 6 (3 terms)
Group n: has n terms.
First, we need to find the first term of the nth group.
Let \( N_k \) be the number of terms in group \( k \). So \( N_k = k \).
The total number of terms before the nth group starts is the sum of terms in groups 1 to \( (n-1) \):
Total terms before group n = \( 1 + 2 + 3 + \dots + (n-1) = \frac{(n-1)n}{2} \).
The last term of group \( (n-1) \) is \( \frac{(n-1)n}{2} \).
So, the first term of the nth group, let's call it \( A_n \), will be one more than the last term of group \( (n-1) \).
\( A_n = \frac{(n-1)n}{2} + 1 \).
\( A_n = \frac{n^2 - n + 2}{2} \).
The last term of the nth group, \( L_n \), is the sum of terms up to group \( n \).
Total terms up to group n = \( 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2} \).
So, \( L_n = \frac{n(n+1)}{2} \).
Now, we need to find the sum of the numbers in the nth group. The nth group consists of n terms, starting from \( A_n \) and ending at \( L_n \), with a common difference of 1 (since they are natural numbers).
The sum of an arithmetic progression is \( S_k = \frac{k}{2} (first \text{ term} + last \text{ term}) \).
Here, \( k=n \) (number of terms in the nth group), first term is \( A_n \), and last term is \( L_n \).
Sum of nth group = \( \frac{n}{2} (A_n + L_n) \).
Substitute the expressions for \( A_n \) and \( L_n \):
Sum of nth group = \( \frac{n}{2} \left( \frac{n^2 - n + 2}{2} + \frac{n(n+1)}{2} \right) \).
Combine the terms inside the parenthesis (they already have a common denominator of 2):
Sum of nth group = \( \frac{n}{2} \left( \frac{n^2 - n + 2 + n^2 + n}{2} \right) \).
Simplify the numerator inside the parenthesis:
Sum of nth group = \( \frac{n}{2} \left( \frac{2n^2 + 2}{2} \right) \).
Sum of nth group = \( \frac{n}{2} \left( n^2 + 1 \right) \).
Thus, the sum of the numbers in the nth group is \( \frac{n}{2}(n^2 + 1) \). This confirms the statement.
In simple words: We first find what the very first number in the nth group is and what the very last number is. We do this by summing up the numbers of terms in all the groups before it. Since each group contains natural numbers in order, it's like an arithmetic series. We then use the formula for the sum of an arithmetic series to find the total sum of all the numbers in that nth group.

🎯 Exam Tip: For problems involving grouped terms, always determine the number of terms in each group first. The first and last terms of a group are found by summing the terms of preceding groups. These problems often combine AP concepts with series summation.

Question 16. If the sum of first n terms of an A.P. is cn², then the sum of squares of these n terms is
(a) \( \frac{n(4n^2-1)c^2}{6} \)
(b) \( \frac{n(4n^2+1)c^3}{3} \)
(c) \( \frac{n(4n^2-1)c^2}{3} \)
(d) \( \frac{n(4n^2+1)c^2}{6} \)
Answer: (c) \( \frac{n(4n^2-1)c^2}{3} \)
To find the sum of squares of the terms of the AP, we first need to find the general term \( T_n \) of the AP.
Given that the sum of the first n terms of an A.P. is \( S_n = cn^2 \).
The nth term \( T_n \) can be found using the formula \( T_n = S_n - S_{n-1} \).
\( S_{n-1} = c(n-1)^2 \).
So, \( T_n = cn^2 - c(n-1)^2 \).
\( T_n = c [n^2 - (n-1)^2] \).
Expand \( (n-1)^2 = n^2 - 2n + 1 \).
\( T_n = c [n^2 - (n^2 - 2n + 1)] \).
\( T_n = c [n^2 - n^2 + 2n - 1] \).
\( T_n = c(2n - 1) \). This is the nth term of the arithmetic progression.
Now, we need to find the sum of the squares of these n terms, which is \( \sum T_n^2 \).
\( T_n^2 = [c(2n - 1)]^2 = c^2 (2n - 1)^2 \).
Expand \( (2n-1)^2 = (2n)^2 - 2(2n)(1) + 1^2 = 4n^2 - 4n + 1 \).
So, \( T_n^2 = c^2 (4n^2 - 4n + 1) \).
Now, we need to sum this expression for n terms:
\( \sum T_n^2 = \sum c^2 (4n^2 - 4n + 1) \).
\( \sum T_n^2 = c^2 \sum (4n^2 - 4n + 1) \).
\( \sum T_n^2 = c^2 \left( 4 \sum n^2 - 4 \sum n + \sum 1 \right) \).
Substitute the standard sum formulas:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
\( \sum 1 = n \)
\( \sum T_n^2 = c^2 \left[ 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n \right] \).
Simplify the expression:
\( \sum T_n^2 = c^2 \left[ \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \right] \).
Take out the common factor \( \frac{n}{3} \) from inside the square bracket (to make a common denominator of 3):
\( \sum T_n^2 = c^2 \frac{n}{3} \left[ 2(n+1)(2n+1) - 6(n+1) + 3 \right] \).
Simplify the terms inside the square bracket:
\( \sum T_n^2 = c^2 \frac{n}{3} \left[ 2(2n^2 + 3n + 1) - (6n + 6) + 3 \right] \).
\( \sum T_n^2 = c^2 \frac{n}{3} [ 4n^2 + 6n + 2 - 6n - 6 + 3 ] \).
\( \sum T_n^2 = c^2 \frac{n}{3} [ 4n^2 - 1 ] \).
\( \sum T_n^2 = \frac{n(4n^2 - 1)c^2}{3} \). This matches option (c). A sum of squares can often be simplified to a compact form.
In simple words: First, we use the formula for the sum of 'n' terms to find the actual formula for the nth term of the arithmetic progression. Then, we square this nth term and sum up these squared terms. We use standard sum formulas for powers of 'n' to get the final answer.

🎯 Exam Tip: To find the nth term of a series when only the sum to n terms (\( S_n \)) is given, use the formula \( T_n = S_n - S_{n-1} \). This is a very common and important technique. Remember to expand and factor carefully.