Find the sum to n terms of the series 3 x 1² + 5 x 2² + 7 x 3² + ....

Solution:

The given series is 3 x 12 + 5 x 22 + 7 x 32 + .... n terms.

Hence,

a = (2n + 1) n²

= 2n³ + n²

Therefore,

S_n = ∑ⁿ_{k = i}(a)_k

= ∑ⁿ_{k = 1}(2k³ + k²)

= 2∑ⁿ_{k = 1}(k)³ + ∑ⁿ_{k = 1}(k)²

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

= n² (n + 1)²/2 + [n (n + 1)(2n + 1)]/6

= n (n + 1)/2 [n (n + 1) + (2n + 1)/3]

= n (n + 1)/2 [3n² + 3n + 2n + 1]/3

= n (n + 1)/2 [3n² + 5n + 1]/3

= n/6 (n + 1)(3n² + 5n + 1)

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 3

Find the sum to n terms of the series 3 x 1² + 5 x 2² + 7 x 3² + ....

Summary:

We found out using the sum to n terms of the series 3 x 1² + 5 x 2² + 7 x 3² + .... up to n terms is n/6 (n + 1)(3n² + 5n + 1) using a = (2n + 1) n²