# Find the sum to n terms of the series 3 x 1^{2} + 5 x 2^{2} + 7 x 3^{2} + ....

**Solution:**

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

Hence,

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

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

Therefore,

S_{n} = ∑^{n}_{k = i}(a)_{k}

= ∑^{n}_{k = 1}(2k^{3} + k^{2})

= 2∑^{n}_{k = 1}(k)^{3} + ∑^{n}_{k = 1}(k)^{2}

= 2[n (n + 1)/2]^{2} + [n (n + 1)(2n + 1)]/6

= n^{2} (n + 1)^{2}/2 + [n (n + 1)(2n + 1)]/6

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

= n (n + 1)/2 [3n^{2} + 3n + 2n + 1]/3

= n (n + 1)/2 [3n^{2} + 5n + 1]/3

= n/6 (n + 1)(3n^{2} + 5n + 1)

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 3

## Find the sum to n terms of the series 3 x 1^{2} + 5 x 2^{2} + 7 x 3^{2} + ....

**Summary:**

We found out using the sum to n terms of the series 3 x 1^{2} + 5 x 2^{2} + 7 x 3^{2} + .... up to n terms is n/6 (n + 1)(3n^{2} + 5n + 1) using a = (2n + 1) n^{2}

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