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

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4

Find the sum of n terms of the following series:
1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ……….
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 ×4 × 5 + ……….
3. 3 × 1² + 5 × 2² + 7 × 3² + ………
4. $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ + …………….
5. 5² + 6² + 7² + ……………. + 20²
6. 3 × 8 + 6 × 11 + 9 × 14 + ……………….
7. 1² + (1² + 2²) + (1² + 2² + 3²) + ………………..
Solutions to Q.No. 1 – 7:
1. Let S = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
now, nth term = n × (n + 1)
⇒ T_n = n² + n.

2. Let T_n denote the nth term of the given series. Then,
T_n = [nth term of 1, 2, 3,…] [nth term of 2, 3, 4,…] [nth term of 3, 4, 5, …]
= [1 + (n – 1)1][2 + (n – 1). 1][3 + (n – 1).1]
= n(n + 1)(n + 2) = n(n² + 2n + 3)
= n³ + 3n² + 2n.
S_n = Σn³ + 3Σn² + 2Σn.

3. Let T_n denote the nth term of the given series. Then,
T_n = [nth term of 3, 5, 7, …………..][nth term of 1, 2, 3, …………….]²
= [3 + 2(n – 1)²3[1 + (n – 1).1²] = (2n + 1)(n)
= n²(2n + 1) = 2n³ + n².
So, S_n = 2Σn³ + Σn².

4.

Let $\frac{1}{n(n+1)}$ = $\frac{A}{n}$ + $\frac{B}{n+1}$ …………….. (1)
⇒ 1 = A(n + 1) + Bn …………………. (2) [On multiplying both sides by n(n + 1)]

To find B:
Put n = – 1 in (2), we get
1 = B(- 1) ⇒ B = – 1.
Put these values of A and B in (1), the partial fractions are
$\frac{1}{n(n+1)}$ = $\frac{1}{n}$ – $\frac{1}{n+1}$.
∴ T_n = $\frac{1}{n}$ – $\frac{1}{n+1}$.
Putting n = 1, 2, 3, ………….., n, we get

Adding vertically, we get
S_n = 1 – $\frac{1}{n+1}$ = $\frac{n+1-1}{n+1}$ = $\frac{n}{n+1}$.

5. T_n of the given series
= (nth term of 5, 6, 7, ………….)²
= [5 + (n – 1).1]²
= (n + 4)² = n² + 8n + 16.
∴ S_n = Σn² + 8Σn + 16 × n.
= $\frac{n(n+1)(2n+1)}{6}$ + 8 × $\frac{n(n+1)}{2}$ + 16n
= $\frac{n}{6}$[(n + 1)(2n + 1) + 24(n + 1) + 96]
= $\frac{n}{6}$[2n² + n + 2n + 1 + 24n + 24 + 96]
= $\frac{n}{6}$(2n² + 27n + 121).
Put n = 16, then
S = $\frac{16}{6}$ (2 × 16² + 27 × 16 + 121)
= $\frac{8}{3}$ (512 + 432 + 121) = $\frac{8}{3}$ × 1065
= 8 × 355 = 2840.

6. Here, the series is formed by multiplying the corresponding terms of two series both of which are A.P.
viz., 3, 6, 9,… and 8, 11, 14, …
∴ T_n of given series = (nth term of 3, 6, 9, …) × (nth term of 8, 11, 14, …)
= [3 + (n – 1)3] [8 + (n – 1)3]
= [3n][3n + 5].
So, S_n = 9n² + 15n = 9Σn² + 15Σn.
= 9 × $\frac{n(n+1)(2n+1)}{6}$ + $\frac{15.n(n+1)}{2}$
= $\frac{3}{2}$.n(n + 1)[2n + 1 + 5].
= $\frac{3}{2}$.n(n + 1)(2n + 6) = 3n(n + 1)(n + 3).

7. Let T_n denote the nth term. Then,

Find the sum of first n terms of the following series whose nth terms are given by:
8. n(n + 1)(n + 4)
9. n² + 2ⁿ
10. (2n – 1)²
Solutions to Q.No. 8 – 10:
8. T_n = n(n + 1)(n + 4) = n(n² + 5n + 4)
= n³ + 5n² + 4n.

9. T_n = n² + 2ⁿ
Put n = 1, 2, 3, ………….. n, we get
T₁ = 1² + 2¹
T₂ = 2² + 2²
T₃ = 3² + 2³
……………………
T_n = n² + 2ⁿ
Adding vertically, we have:
S_n = (1² + 2² + 3² + …………… + n²) + (2¹ + 2² + 2³ + …………….. + 2ⁿ)
= Σn² + $\frac{2\left(2^{n}-1\right)}{2-1}$ = $\frac{n(n + 1)(2n + 1)}{6}$ + 2(2ⁿ – 1).

10. T_n = (2n – 1)² = 4n² – 4n + 1