RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4 Textbook Exercise Questions and Answers.

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

Find the sum to n terms of each of the series in exercise 1 to 7

Question 1.
1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + ...................
Answer:
Let the sum of n terms of series be S_n.
S_n = (1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + ......... + n terms
Now,n^th term of given series
= [1, 2, 3, 4, ........ nth term] × n^th term of [2, 3, 4, 5, ...............]
T_n = [1 + (n - 1).1] × [2 + (n - 1).1]
[∵ n^th term of A.P. = a + (n - 1)d]
⇒ T _n = n(n + 1)

Question 2.
1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 ..............
Answer:
Let the sum of n terms of the sequence is S.
S_n = 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + .............. n terms
n^th term of series = n^th term of [1, 2, 3, ..........] × n^th term of [2, 3, 4, ..............] × n term of [3, 4, 5, ................]
T_n = [1 + (n - 1).1] × [2 + (n - 1).1] × [3 + (n - 1).1]
[∵ n^th term of A.P. = 9 + (n - 1)d]
T_n = (1 + n - 1) (2 + n - 1) (3 + n - 1)
= n (n + 1) (n + 2)
= n(n² + 3n + 2)
= n³ + 3n² + 2n

∴ S_n = \(\frac{1}{4}\) n(n + 1) (n + 2) (n + 3)
Thus, sum of n terms of the given series
= \(\frac{1}{4}\) n(n + 1) (n + 2) (n + 3)

Question 3.
3 × 1² + 5 × 2² + 7 × 3² + .............
Answer:
Let sum of n terms of series be Sn.
Then S_n = 3 × 1² + 5 × 2² + 7 × 3² + .................. + n terms
n^th term of series = (nth term of 3 + 5 + 7 + .... ) × n²
[∵ a_n = a + (n - 1)d]
= [3 + (n - 1) × 2]n²
= [3 + 2n - 2] × n²
= (2n + 1) n²
= 2n³ + n²

Question 4.
\(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots \ldots\)
Answer:
Let the sum of n terms of series is S_n.

Question 5.
5² + 6² + 7² + .......... + 20²
Answer:
Given series = 5² + 6² + 7² + ....................... + 20²
So, n^th term of series
= [n^th term of sequence]
= [5 + (n - 1) × 1]² (∵ 5, 6, 7, ... are in A.P.)
= [5 + n - 1]²
= (n + 4)² = n² + 8n + 16
Sum of n terms of series

We see that last term of sequence 5, 6, 7, ........ 20 is 20.
Since, T_n = a + (n - 1)d
20 = 5 + (n - 1) × 1
⇒ 20 = 5 + n - 1
⇒ 20 = n + 4
⇒ n = 16
Putting n =16 in equation (1), we have
S₁₆ = \(\frac{16(16+1)(2 \times 16+25)}{6}\) + 16 × 16
= \(\frac{16 \times 17 \times 57}{6}\) + 256
= 8 × 17 × 19 + 256
= 2584 + 256
= 2840
Thus, 5² + 6² + 7² + ...................... + 20² = 2840

Note: Since, series is 5² + 6² + 7² + ...................... + 20² and square of numbers are from 5 to 20. Thus, number of terms will be 16.

Question 6.
3× 8 + 6 × 11 + 9 × 14 + ...........
Answer:
Given series 3 × 8 + 6 × 11 + 9 × 14 + .............. n terms
Let n^th term of sequence be T_n.
T_n = (n^th term of sequence 3, 6, 9, ............ ) × (n^th term of sequence 8, 11, 14 ............)
= [3 + (n - 1) × 3] × [8 + (n - 1) × 3]
[∵ n^th term of A.p. = a + (n - 1)d]
= [3 + 3n - 3] × [8 + 3n - 3]
= 3n × (3n + 5)
= 9n² + 15n
Let S_n be the sum of n term of the series.

= n(n + 1) (3n + 9)
= 3n (n + 1) (n + 3)
Thus, sum of n terms of the given series
S_n = 3n(n + 1) (n + 3)

Question 7.
1² + (1² + 2²) + (1² + 2² + 3²) + ..............
Answer:
Given series = 1² + (1² + 2²) + (1² + 2² + 3²) + .............. n terms
n^th term of series = 1² + 2² + ............... + n²

Find the sum of n terms of the series ¡n exercIse 8 to 10 whose n^th term is given by:

Question 8.
n(n + 1) (n + 4)
Answer:
nth term of the given series = n (n + 1) (n + 4)
T_n = n(n² + 5n + 4)
T_n = n³ + 5n² + 4n
Let S_n be the sum of n terms of the sequence.

Question 9.
n² + 2ⁿ
Answer:
n^th term of the series T_n = n² + 2ⁿ

Question 10.
(2n - 1)²
Answer:
n^th term of given series T_n = (2n - 1)²
= 4n² - 4n + 1
Let S_n be the sum of n terms of the sequence