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)
⇒ Tn = n2 + n.
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 1

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

2. Let Tn denote the nth term of the given series. Then,
Tn = [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.
Sn = Σn3 + 3Σn2 + 2Σn.
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 2

3. Let Tn denote the nth term of the given series. Then,
Tn = [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, Sn = 2Σn3 + Σn2.
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 3

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

4.
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 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
\(\frac{1}{n(n+1)}\) = \(\frac{1}{n}\) – \(\frac{1}{n+1}\).
∴ Tn = \(\frac{1}{n}\) – \(\frac{1}{n+1}\).
Putting n = 1, 2, 3, ………….., n, we get
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 5
Adding vertically, we get
Sn = 1 – \(\frac{1}{n+1}\) = \(\frac{n+1-1}{n+1}\) = \(\frac{n}{n+1}\).

5. Tn of the given series
= (nth term of 5, 6, 7, ………….)2
= [5 + (n – 1).1]2
= (n + 4)2 = n2 + 8n + 16.
∴ Sn = 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.

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

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, …
∴ Tn 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, Sn = 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 Tn denote the nth term. Then,
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 6

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. Tn = n(n + 1)(n + 4) = n(n2 + 5n + 4)
= n3 + 5n2 + 4n.
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 7

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

9. Tn = n2 + 2n
Put n = 1, 2, 3, ………….. n, we get
T1 = 12 + 21
T2 = 22 + 22
T3 = 32 + 23
……………………
Tn = n2 + 2n
Adding vertically, we have:
Sn = (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).

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

10. Tn = (2n – 1)2 = 4n2 – 4n + 1
GSEB Solutions Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 img 8

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