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 × 1^{2} + 5 × 2^{2} + 7 × 3^{2} + ………

4. \(\frac{1}{1×2}\) + \(\frac{1}{2×3}\) + \(\frac{1}{3×4}\) + …………….

5. 5^{2} + 6^{2} + 7^{2} + ……………. + 20^{2}

6. 3 × 8 + 6 × 11 + 9 × 14 + ……………….

7. 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ………………..

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} = n^{2} + 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(n^{2} + 2n + 3)

= n^{3} + 3n^{2} + 2n.

S_{n} = Σn^{3} + 3Σn^{2} + 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)^{2}3[1 + (n – 1).1^{2}] = (2n + 1)(n)

= n^{2}(2n + 1) = 2n^{3} + n^{2}.

So, S_{n} = 2Σn^{3} + Σn^{2}.

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} = n^{2} + 8n + 16.

∴ S_{n} = Σn^{2} + 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]

= \(\frac{n}{6}\)[2n^{2} + n + 2n + 1 + 24n + 24 + 96]

= \(\frac{n}{6}\)(2n^{2} + 27n + 121).

Put n = 16, then

S = \(\frac{16}{6}\) (2 × 16^{2} + 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} = 9n^{2} + 15n = 9Σn^{2} + 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. n^{2} + 2^{n}

10. (2n – 1)^{2}

Solutions to Q.No. 8 – 10:

8. T_{n} = n(n + 1)(n + 4) = n(n^{2} + 5n + 4)

= n^{3} + 5n^{2} + 4n.

9. T_{n} = n^{2} + 2^{n}

Put n = 1, 2, 3, ………….. n, we get

T_{1} = 1^{2} + 2^{1}

T_{2} = 2^{2} + 2^{2}

T_{3} = 3^{2} + 2^{3}

……………………

T_{n} = n^{2} + 2^{n}

Adding vertically, we have:

S_{n} = (1^{2} + 2^{2} + 3^{2} + …………… + n^{2}) + (2^{1} + 2^{2} + 2^{3} + …………….. + 2^{n})

= Σn^{2} + \(\frac{2\left(2^{n}-1\right)}{2-1}\) = \(\frac{n(n + 1)(2n + 1)}{6}\) + 2(2^{n} – 1).

10. T_{n} = (2n – 1)^{2} = 4n^{2} – 4n + 1