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# Given $\;t_{r}=1^2+2^2+.......\;r^2$ and $t_{1}+t_{2}+t_{3}+...\;t_{n}=\large\frac{k}{12}\;n\;(n+1)\;(n+2)\;$ the value of k will be

$(a)\;n\qquad(b)\;2n+1\qquad(c)\;3n-1\qquad(d)\;n+1$

Answer : (d) $\;n+1$
Explanation : $t_{r}=\large\frac{r(r+1)(2r+1)}{6}$
$\sum\;t_{r}=\sum_{r=1}^{n}\;\large\frac{r(r+1)(2r+1)}{6}$
$=\sum_{r=1}^{n}\;[\large\frac{r(r+1)(2r+1)}{3}-\large\frac{1}{2}\;r(r+1)]$
$=\large\frac{1}{12}\;n(n+1)(n+2)(n+3)-\large\frac{1}{6}\;n(n+1)(n+2)$
$=\large\frac{1}{6}\;n(n+1)(n+2)\;[\large\frac{n+3}{2}-1]$
$=\large\frac{1}{12}\;n(n+1)^2(n+2).$