## 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) 3n^{2} + 2n + 5,

(ii) 4n^{3} + 7n + 1

(iii) n(n + 1)(n + 2)

Solution:

(iii) Here T_{n} = n (n + 1)(n + 2)

⇒ T_{n} = n(n^{2} + 3n + 2)

⇒ T_{n} = n^{3} + 3n^{2} + 2n

∴ S_{n} = ∑T_{n}

⇒ S_{n} = ∑n^{3} + 3∑n^{2} + 2∑M

Question 2.

Find the sum of it terms of following series :

(i) 3^{2} + 7^{2} + 11^{2} + 15^{2} + …

(ii) 2^{3} + 5^{3} + 8^{3} + 11^{3} + …

(iii) 1.2^{2} + 2.3^{2} + 3.4^{2} + …

Solution:

(i) 3^{2} + 7^{2} + 11^{2} + 15^{2} + …

n^{th} term of the given series

T_{n} = [3 + (n – 1)4]^{2}

= (3 + 4n – 4)^{2}

= (4n – 1 )^{2} = 16n^{2} + 1 – 8n

∴ S_{n} = ∑(16n^{2} – 8n + 1)

⇒ Sn = 16∑n^{2} – 8∑M + ∑1

(ii) 2^{3} + 5^{3} + 8^{3} + 11^{3} + …

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) × 1^{2} – 1^{3}

= 27n^{3} – 27n^{2} + 9n – 1

∴ S_{n} = ∑(27n^{3} – 27n^{2} + 9n – 1)

⇒ s_{n} = 27∑n^{3} – 27∑n^{2} + 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(n^{2} + 2n + 1)]

= ∑[n^{3} + 2 n^{2} + n]

= ∑n^{3} + 2∑M^{2} + ∑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} = ∑ (4n^{2} – 1)

= 4∑n^{2 }– ∑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)

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

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

= 3n^{3} + 4n^{2} + n = n(n + 1) (3n + 1)

∴ Sn = ∑(3n^{3} + 4n^{2} + n)

= 3∑n^{3} + 4∑n^{2} + ∑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) 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + …

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

Let n^{th} term of the series be T_{n} and sum of n terms is s_{n}, then

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