AP Inter 1st Year Maths Exercise 8e Solutions

Referring to the AP Inter 1st Year Maths Study Material Chapter 8 Sequences and Series Exercise 8e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Sequences and Series Solutions Exercise 8e

I. Find the sum to n terms of each of the series in Exercises 1 to 7.

Question 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Solution:
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … n terms
Hence, a_n = n(n + 1)
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k(k + l) = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+1}{3}+1\right)\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\frac{2 \mathrm{n}+4}{3}\right)\)

= \(\frac{\mathrm{n}}{3}\)(n + 1)(n + 2)

Question 2.
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Solution:
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + … n terms
Hence, a_n = n(n + 1) (n + 2) = (n² + n) (n + 2) = n³ + 3n² + 2n
Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\)(k³ + 3k² + 2k) = \(\sum_{k=1}^n\) k³ + 3 \(\sum_{k=1}^n\) k² + 2 \(\sum_{k=1}^n\) k

Question 3.
3 × 1² + 5 × 2² + 7 × 3² + …
Solution:
The given series is 3 × 1² + 5 × 2² + 7 × 3² + …. n terms
Hence, a_n = (2n + 1) n² = 2n³ + n²
Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a_k = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (2k³ + k²) = 2\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k³ + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k²

= 2\(\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\) + \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{2}+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

= \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left(\mathrm{n}(\mathrm{n}+1)+\frac{2 \mathrm{n}+1}{3}\right)\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+3 n+2 n+1}{3}\right]\)

= \(\frac{n(n+1)}{2}\left[\frac{3 n^2+5 n+1}{3}\right]\)

= \(\frac{n}{6}\)(n + 1)(3n + 5n + 1)

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ………..
Solution:
The given series is \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}\) + ……….. n terms
Hence, a_n = \(\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
∴ a₁ = \(\frac{1}{1}-\frac{1}{2}\), a₂ = \(\frac{1}{2}-\frac{1}{3}\), a₃ = \(\frac{1}{3}-\frac{1}{4}\), …………….., a_n = \(\frac{1}{n}-\frac{1}{n+1}\),

Adding the above terms column wise, we obtain
a₂ + a₂ + …… + a_n = \(\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{\mathrm{n}}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{\mathrm{n}+1}\right]\)

Thus, S_n = 1 – \(\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)

Question 5.
5² + 6² + 7² + ………. +20²
Solution:
The given series is 5² + 6² + 7² + ………. +20²
Hence, a_n = (n + 4)² = n² + 8n + 16

Therefore, S_n = \(\sum_{k=1}^n\)a_k = \(\sum_{k=1}^n\)(k² + 8k + 16)

S_n = \(\sum_{k=1}^n\)k² + 8\(\sum_{k=1}^n\)k + \(\sum_{k=1}^n\) 16

= \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{8 \mathrm{n}(\mathrm{n}+1)}{2}\) + 16n

Since, 16^th term is (16 + 4)² = (20)²

Then, S_n = \(\frac{16(16+1)(2(16)+1)}{6}+\frac{8(16)(16+1)}{2}\)+ 16(16)

= \(\frac{(16)(17)(33)}{6}+\frac{(128)(17)}{2}\) + 256

= 1496 + 1088 + 256 = 2840

Thus, 5² + 6² + 7² + ………. +20² = 2840

Question 6.
3 × 8 + 6 × 11 + 9 × 14 + …………..
Solution:
The given series is 3 × 8 + 6 × 11 + 9 × 14 + ………….. n terms
Hence, a_n = (n^th term of 3, 6, 9, ………..) × (n^th term of 8, 11, 14 ……)
= (3n)(3n + 5)
= 9n² + 15n

Therefore, S_n = \(\sum_{k=1}^n\) ak = \(\sum_{k=1}^n\) (9k² + 15k)

= 9 \(\sum_{k=1}^n\) k² + 15\(\sum_{k=1}^n\)k

= 9 × \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\) + 15 × \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{2}+\frac{15 \mathrm{n}(\mathrm{n}+1)}{2}\)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 1 + 5)

= \(\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}\)(2n + 6)

= 3n(n + 1)(n + 3)

Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ………..
Solution:
The given series is 1² + (1² + 2²) + (1² + 2² + 3²) + ……….. n terms

Hence, a_n = (1² + 2² + 3² + …… + n²) = \(\frac{n}{2}\)(n + 1)(2n + 1)
= \(\frac{n}{6}\) (2n² + 3n + 1)
= \(\frac{1}{3}\)n³ + \(\frac{1}{2}\)n² + \(\frac{1}{6}\)n

Therefore,

II. Find the sum to n terms of the series in Exercises 1 to 3 whose nth terms is given by

Question 1.
n(n + 1) (n + 4).
Solution:
The given n* term is a_n = n(n + 1) (n + 4)
Hence, a_n = n(n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n
Therefore, S_n = \(\sum_{k=1}^n\) a_k = \(\sum_{k=1}^n\)k³ + 5 \(\sum_{k=1}^n\) k² + 4 \(\sum_{k=1}^n\) k

Question 2.
n² + 2n
Solution:
The given n^th term is an = n² + 2ⁿ
Hence, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + 2^k

= \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)2^k …………..(1)
Consider \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = 2¹ + 2² + 2³ +…

The above series 2¹ + 2² + 2³ + … is a G.P with both the first term and common ratio equal to 2.
\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k = \(\frac{1}{2}\) = 2(2ⁿ – 1) …………..(2)

From (1) and (2) we obtain
S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k² + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 2^k
= \(\frac{n}{6}\)(n + 1)(2n + 1) + 2(2ⁿ – 1)

Question 3.
(2n – 1)²
Solution:
The given n^th term is a_n = (2n – 1)²
Hence, a_n = (2n – 1)² = 4n² – 4n + 1

Therefore, S_n = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) a = \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) (4k² – 4k + 1)

= 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\)k2 – 4\(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) k + \(\sum_{\mathrm{k}=1}^{\mathrm{n}}\) 1

= \(\frac{4 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{4 \mathrm{n}(\mathrm{n}+1)}{2}\) + n

= \(\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{3}\) – 2n(n + 1) + n

= n[\(\frac{2\left(2 n^2+3 n+1\right)}{3}\) – 2(n + 1) + 1]

= n\(\left[\frac{4 n^2+6 n+2-6 n-6+3}{3}\right]\) = n\(\left[\frac{4 \mathrm{n}^2-1}{3}\right]\)

= \(\frac{\mathrm{n}}{3}\) (2n + 1)(2n – 1)