Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

Asked by Pragya Singh | 1 year ago |  138

##### Solution :-

Given series is 12 + (12 + 22) + (12 + 2+ 32 ) + …

Finding the nth term, we have

an = (12 + 22 + 32 +…….+ n2)

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

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

$$\dfrac{2n^2+3n^2+n}{6}$$

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

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

$$S_n= \displaystyle\sum_{k=1}^{n}= (\dfrac{1}{3}k^3+\dfrac{1}{2}k^2+\dfrac{1}{6}k)$$

$$\dfrac{1}{3}\displaystyle\sum_{k=1}^{n}=k^3+\dfrac{1}{2}\displaystyle\sum_{k=1}^{n}k^2+\dfrac{1}{6}\displaystyle\sum_{k=1}^{n}k$$

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

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

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

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

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

$$\dfrac{n(n+1)}{6}[\dfrac{(n+1)+(n+2)}{12}]$$

$$\dfrac{n(n+1)(n+2)}{12}$$

Answered by Pragya Singh | 1 year ago

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