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Q)

Sum of series $(n)(n)$ + $(n-1)(n+1)$ + $(n-2)(n+2)$ +....+ $1\;(2n-1)$ is

$(a)\;n^3\qquad(b)\;\frac{1}{6}\;n\;(n+1)\;(n+2)\qquad(c)\;\frac{1}{3}n^3-n^2\qquad(d)\;None$

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A)
Answer : (d) None
Explanation : $S_{n}=\sum_{r=0}^{r=n-1}\;(n-r)(n+r)=\sum\;n^2-r^2$
$=n^3-\frac{1}{6}\;n\;(n-1)\;(2n-1).$
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