The sum 1+(1)/(2)(1^2+2^2)+(1)/(3)(1^2+2^2+3^2)+ upto 10 terms is equal to :

Question

The sum $1+\frac{1}{2}({1}^{2}+{2}^{2})+\frac{1}{3}({1}^{2}+{2}^{2}+{3}^{2})+\ldots$ upto 10 terms is equal to :

Accepted Answer

$\begin{gathered} {T}_{n}=\frac{1}{n}({1}^{2}+{2}^{2}+\ldots +{n}^{2}) \ {T}_{n}=\frac{n(n+1)(2n+1)}{6n} \ {T}_{n}=\frac{1}{6}[2{n}^{2}+3n+1] \ {S}_{n}=\Sigma {T}_{n}=\frac{1}{6}[2\Sigma {n}^{2}+3\Sigma n+\Sigma 1] \ \Sigma {T}_{n}=\frac{1}{6}[2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{3n(n+1)}{2}+n] \ {S}_{10}=\frac{1}{6}[\frac{2	imes 10	imes 11	imes 21}{6}+\frac{3}{2}	imes 10	imes 11+10] \ =\frac{10}{6}[77+\frac{33}{2}+1]=\frac{315}{2} \end{gathered}$

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