RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.6

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.6

Question 1.
Find the sum of n term of that series whose nth term is :
(i) 3n² + 2n + 5,
(ii) 4n³ + 7n + 1
(iii) n(n + 1)(n + 2)
Solution:


(iii) Here T_n = n (n + 1)(n + 2)
⇒ T_n = n(n² + 3n + 2)
⇒ T_n = n³ + 3n² + 2n
∴ S_n = ∑T_n
⇒ S_n = ∑n³ + 3∑n² + 2∑M

Question 2.
Find the sum of it terms of following series :
(i) 3² + 7² + 11² + 15² + …
(ii) 2³ + 5³ + 8³ + 11³ + …
(iii) 1.2² + 2.3² + 3.4² + …
Solution:
(i) 3² + 7² + 11² + 15² + …
n^th term of the given series
T_n = [3 + (n – 1)4]²
= (3 + 4n – 4)²
= (4n – 1 )² = 16n² + 1 – 8n
∴ S_n = ∑(16n² – 8n + 1)
⇒ Sn = 16∑n² – 8∑M + ∑1

(ii) 2³ + 5³ + 8³ + 11³ + …
n^th term of the given series
T_n = [2 + (n – 1)3]³
= (2 + 3M – 3)³ = (3n – 1)³
= (3n)³ – 3(3n)² × 1 + 3(3n) × 1² – 1³
= 27n³ – 27n² + 9n – 1
∴ S_n = ∑(27n³ – 27n² + 9n – 1)
⇒ s_n = 27∑n³ – 27∑n² + 9∑n – ∑1

(iii) 1.22 + 2.32 + 3.42 + …
n^th term of the given series
T_n = n.(n + 1)²
∴ S_n = ∑n(n + 1 )²
⇒ S_n = ∑[n(n² + 2n + 1)]
= ∑[n³ + 2 n² + n]
= ∑n³ + 2∑M² + ∑n

Question 3.
Find the n^th term and sum of n terms of following series :
(i) 1.3+3.5+ 5.7 + …
(ii) 1.2.4 + 2.3.7 + 3.4.10 + …
Solution:
(i) 1.3 + 3.5 + 5.7 + …
n^th term of the given series
T_n = (1 + 3 + 5 + … n^th term)
(3 + 5 + 7 + … n^th term)
= [1 + (n – 1)² [3 +(n – 1)2]
= (1 + 2n – 2) (3 + 2n – 2)
= (2n – 1) (2n + 1)
∴ S_n = ∑ (4n² – 1)
= 4∑n² – ∑1

(ii) 1.2.4 + 2.3.7 + 3.4.10 + …
n^th term of the given series
T_n = (1 + 2 + 3 + … n^th term)
(2 + 3 + 4 + … n^th term )
(4 + 7 + 10 + … n^th term)
= n . (n + 1) . [4 + (n – 1)³]
= n(n + 1) (3n + 1)
= (n² + n) (3n + 1)
= 3n³ + 3n² + n² + n
= 3n³ + 4n² + n = n(n + 1) (3n + 1)
∴ Sn = ∑(3n³ + 4n² + n)
= 3∑n³ + 4∑n² + ∑M

Question 4.
Find the n^th term and sum of n terms of the following series :
(i) 3 + 8 + 15 + 24 + …
(ii) 1 + 6 + 13 + 22 …
Solution:
(i) 3 + 8 + 15 + 24 + …
Let n^th term of the series is T_n and sum of n terms is s_n then
s_n = 3 + 8 + 15 + 24 + … + T_n
Again s_n = 3 + 8 + 15 + 24 + … + T_{n – 1} + T_n
On subtracting
0 = 3 + 5 + 7 + 9 + … n upto n terms – T_n
⇒ T_n = 3 + 5 + 7 + 9 + … n upto n terms

(ii) 1 + 6 + 13 + 22 +…
The difference of consecutive terms of given series 5, 7, 9, … are in A.P. So n^th term and sum of its n terms will be find by difference method.
Let n^th term of the series be Tn and sum of nth terms is S_n, then
S_n = 1 + 6 + 13 + 22 + … + T_n …(i)
By increasing one place
S_n = 1 + 6 + 13 + … + T_n-1 + T_n …(ii)
Subtracting (ii) from (i)

Question 5.
Find the nth term and sum of n term of the following series :
(i) 1 + (1 + 2) + (1 + 2 + 3) + …
(ii) 1² + (1² + 2²) + (1² + 2² + 3²) + …
Solution:
(i) 1 + (1 + 2) + (1 + 2 + 3) + …
Let n^th term of the series be Tn and sum of n terms is S_n, then

(ii) 1² + (1² + 2²) + (1² + 2² + 3²) + …
Let n^th term of the series be T_n and sum of n terms is s_n, then

RBSE Solutions for Class 11 Maths