Find the sum to n terms of the series 52 + 62 + 72 + … + 202

Asked by Abhisek | 1 year ago |  107

##### Solution :-

Given series is 52 + 62 + 72 + … + 202

It’s seen that,

nth term, an = ( n + 4)2 = n2 + 8n + 16

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

The sum of n terms of a series is given by the equation

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

The sum of n terms of the given series is

$$S_n= \displaystyle\sum_{k=1}^{n}(k^2+8k+16)$$

$$\displaystyle\sum_{k=1}^{n}k^2+8 \displaystyle\sum_{k=1}^{n}k+16$$

$$\dfrac{n(n+1)(2n+1)}{6}+ \dfrac{8n(n+1)}{2}+16n$$

Then,

$$(20)^2=(16+4)^2$$ is the 16th term

Therefore,

$$s_{16} = \dfrac{16(16+1)(2\times 16+1)}{6}+$$

$$\dfrac{8\times 16(16+1)}{2}+16\times 16$$

$$\dfrac{16(17)(33)}{6}+ \dfrac{(8)(16)(17)}{2}+256$$

= 1496 + 1088 + 256

= 2840

Answered by Pragya Singh | 1 year ago

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