Question

# Find the sum of  $$1^{2}+(1^{2}+2^{2})+(1^{2}+2^{2}+3^{2})+.......$$

Solution

## First term$$={1}^{2}$$Second term$$={1}^{2}+{2}^{2}$$Third term$$={1}^{2}+{2}^{2}+{3}^{2}$$...$${n}^{th}$$ term$$={1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}$$Here $${a}_{n}={1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}$$$$=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$$$$=\dfrac{2{n}^{3}+{n}^{2}+2{n}^{2}+n}{6}$$$$=\dfrac{2{n}^{3}+3{n}^{2}+n}{6}$$Now, the sum of $$n$$ terms is $${S}_{n}=\sum_{n=1}^{n}{{a}_{n}}$$$$=\sum_{n=1}^{n}{\dfrac{2{n}^{3}+3{n}^{2}+n}{6}}$$$$=\dfrac{1}{6}\left(2\sum_{n=1}^{n}{{n}^{3}}+3\sum_{n=1}^{n}{{n}^{2}}+\sum_{n=1}^{n}{n}\right)$$$$=\dfrac{1}{6}\left(\dfrac{2{n}^{2}{\left(n+1\right)}^{2}}{2}+\dfrac{3n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{n\left(n+1\right)}{2}\right)$$$$=\dfrac{n\left(n+1\right)}{12}\left[n\left(n+1\right)+\left(2n+1\right)+1\right]$$$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+n+2n+1+1\right]$$$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+3n+2\right]$$$$=\dfrac{n\left(n+1\right)}{12}\left[{n}^{2}+2n+n+2\right]$$$$=\dfrac{n\left(n+1\right)}{12}\left[\left(n+1\right)\left(n+2\right)\right]$$$$=\dfrac{n{\left(n+1\right)}^{2}\left(n+2\right)}{12}$$Thus, the required sum is $$\dfrac{n{\left(n+1\right)}^{2}\left(n+2\right)}{12}$$Mathematics

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