Find the sum of the following series up to n terms: 1³/1 + (1³ + 2³)/(1 + 3) + (1³ + 2³ + 3³)/(1 + 3 + 5) + ....

Solution:

The n^th term of the given series 1³/1 + (1³ + 2³)/(1 + 3) + (1³ + 2³ + 3³)/(1 + 3 + 5) + .... is

a_n = (1³ + 2³ + 3³ + .... n³) / [1 + 3 + 5 + .... + (2n - 1)]

= [n (n +1)/2]² / [1 + 3 + 5 + .... + (2n - 1)] (Using the summation formulas)

Here, 1, 3, 5, .... (2n - 1) is an A.P. with first term a , common difference 2, last term (2n - 1) and number of terms n.

Therefore,

1 + 3 + 5 + .... + (2n - 1) = n/2 [2 x 1 + (n - 1)2] = n²

Hence,

a_n = n² (n + 1)²/4n²

= (n² + 2n + 1)/4

= 1/4 (n²) + 1/2 (n) + 1/4

Thus,

S_n = ∑ⁿₖ ₌ ₁(a)_k

= ∑ⁿₖ ₌ ₁(1/4 (k²) + 1/2 (k) + 1/4)

= 1/4 [n (n + 1)(2n + 1)/6] + 1/2 [n(n + 1)/2] + 1/4 x n

= n [(n + 1)(2n + 1) + 6 (n + 1) + 6]/24

= n [2n² + 3n + 1 + 6n + 6 + 6]/24

= n/24 (2n² + 9n + 13)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 25

Find the sum of the following series up to n terms: 1³/1 + (1³ + 2³)/(1 + 3) + (1³ + 2³ + 3³)/(1 + 3 + 5) + ....

Summary:

The sum of the following series 1³/1 + (1³ + 2³)/(1 + 3) + (1³ + 2³ + 3³)/(1 + 3 + 5) + .... up to n terms is n/24 (2n² + 9n + 13)