Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …

Asked by Abhisek | 1 year ago |  100

##### Solution :-

Given series is 3 ×12 + 5 × 22 + 7 × 32 + …

It’s seen that,

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Then, the sum of n terms of the series can be expressed as

$$S_n= \displaystyle\sum_{k=1}^{n}a_k$$

The sum of n terms of the given series is

$$S_n= \displaystyle\sum_{k=1}^{n}(2k^3+k^2)$$

$$2\displaystyle\sum_{k=1}^{n}k^3-\displaystyle\sum_{k=1}^{n}k^2$$

$$2[\dfrac{n(n+1)}{2}]^2+\dfrac{n(n+1)(2n+1)}{6}$$

$$\dfrac{n^2(n+1)^2}{2}+\dfrac{n(n+1)(2n+1)}{6}$$

$$\dfrac{n(n+1)}{2} [n(n+1)\dfrac{2n+1}{3}]$$

$$\dfrac{n(n+1)}{2}[ \dfrac{3n^2+3n+2n+1}{3}]$$

$$\dfrac{n(n+1)}{2} [ \dfrac{3n^2+5n+1}{3}]$$

$$\dfrac{n(n+1)(3n^2+5n+1)}{6}$$

Therefore, the sum of n terms of the series is $$\dfrac{n(n+1)(3n^2+5n+1)}{6}$$

Answered by Pragya Singh | 1 year ago

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