Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Asked by Abhisek | 1 year ago |  72

##### Solution :-

nth term of the series as:

an = n2 + 2n

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

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

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

Let $$\displaystyle\sum_{k=1}^{n}2^k=2^1+2^2+2^3+...$$

Both the first term and the common ratio of $$2^1+2^2+2^3+...$$ which forms a G.P. is 2 .

Therefore,

$$\displaystyle\sum_{k=1}^{n}2^k=\dfrac{(2)[(2)^n-1]}{2-1}$$

$$2 (2^n-1)$$

Then,

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

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

Therefore, the sum of n terms of the series whose ntg term is given by $$n^2+2^n$$

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

Answered by Pragya Singh | 1 year ago

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