1.22+2.32+3.42+……+n(n+1)2=

1·22+2·32+3·42+......+n(n+1)2 Sn=∑tn =∑n(n+1)2 =∑nn2+2n+1 =∑n3+2n2+n =∑n3+2∑n2+∑n =n2(n+1)24+2n(n+1) (2n+1)6+n(n+1)2 =n(n+1)2 n(n+1)2+4n+23+1 =n(n+1)2 3n2+3n+8n+4+66 =n(n+1) 3n2+11n+1012 =n(n+1) (n+2) (3n+5)12

1.22+2.32+3.42+……+n(n+1)2=

1·22+2·32+3·42+......+n(n+1)2 Sn=∑tn =∑n(n+1)2 =∑nn2+2n+1 =∑n3+2n2+n =∑n3+2∑n2+∑n =n2(n+1)24+2n(n+1) (2n+1)6+n(n+1)2 =n(n+1)2 n(n+1)2+4n+23+1 =n(n+1)2 3n2+3n+8n+4+66 =n(n+1) 3n2+11n+1012 =n(n+1) (n+2) (3n+5)12

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$1 . 2^{2} + 2 . 3^{2} + 3 . 4^{2} + \dots \dots + n {(n + 1)}^{2} =$

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$\frac{n (n + 1) (n + 2) (3 n + 1)}{12}$

$\frac{n (n + 1) (n + 2) (n + 3)}{4}$

$2 n (n + 1) (n + 2) (n + 3)$

$\frac{n (n + 1) (n + 2) (3 n + 5)}{12}$

answer is A.

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Detailed Solution

$\begin{array}{rcl} 1 \cdot 2^{2} + 2 \cdot 3^{2} + 3 \cdot 4^{2} + . . . . . . + n {(n + 1)}^{2} \\ S_{n} & = & \sum t_{n} \\ = & \sum n {(n + 1)}^{2} \\ = & \sum n (n^{2} + 2 n + 1) \\ = & \sum (n^{3} + 2 n^{2} + n) \\ = & \sum n^{3} + 2 \sum n^{2} + \sum n \\ = & \frac{n^{2} {(n + 1)}^{2}}{4} + 2 \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} \\ = & \frac{n (n + 1)}{2} [\frac{n (n + 1)}{2} + \frac{4 n + 2}{3} + 1] \\ = & \frac{n (n + 1)}{2} [\frac{3 n^{2} + 3 n + 8 n + 4 + 6}{6}] \\ = & \frac{n (n + 1) (3 n^{2} + 11 n + 10)}{12} \\ = & \frac{n (n + 1) (n + 2) (3 n + 5)}{12} \end{array}$