Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Asked by Abhisek | 1 year ago |  87

##### Solution :-

Given,

nth term of the series as:

an = (2n – 1)2 = 4n2 – 4n + 1

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 first n terms of the given series is

$$S_n= \displaystyle\sum_{k=1}^{n}(4k^2-4k+1)$$

$$= 4\displaystyle\sum_{k=1}^{n}k^2-4\displaystyle\sum_{k=1}^{n}k+4\displaystyle\sum_{k=1}^{n}1$$

$$\dfrac{4n(n+1)(2n+1)}{6}-$$

$$\dfrac{4n(n+1)}{2}+n$$

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

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

$$n[ \dfrac{4n^2+6n+2-6n-6+3}{3}]$$

$$n[ \dfrac{4n^2+1}{3}]$$

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

Therefore, the sum of n terms of the series whose nth term is given by(2n-1)2 is

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

Answered by Pragya Singh | 1 year ago

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