12.221+22.3213+23+32.4213+23+33+……n terms =

tn=n2·n+12n2(n+1)24=1n(n+1)tn=1n-1n+1Sn=∑1n 1n-1n+1 =1-12+12-13+.......+1n-1n+1 =nn+1

12.221+22.3213+23+32.4213+23+33+……n terms =

tn=n2·n+12n2(n+1)24=1n(n+1)tn=1n-1n+1Sn=∑1n 1n-1n+1 =1-12+12-13+.......+1n-1n+1 =nn+1

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$\frac{\frac{1}{2} . \frac{2}{2}}{1} + \frac{\frac{2}{2} . \frac{3}{2}}{1^{3} + 2^{3}} + \frac{\frac{3}{2} . \frac{4}{2}}{1^{3} + 2^{3} + 3^{3}} + \dots \dots n$ terms =

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$\frac{n^{2}}{{(n + 1)}^{2}}$

$\frac{n^{3}}{{(n + 1)}^{3}}$

$\frac{n}{n + 1}$

$\frac{1}{n + 1}$

answer is C.

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Detailed Solution

$\begin{array}{rcl} t_{n} & = & \frac{\frac{n}{2} \cdot \frac{n + 1}{2}}{\frac{n^{2} {(n + 1)}^{2}}{4}} = \frac{1}{n (n + 1)} \\ t_{n} & = & \frac{1}{n} - \frac{1}{n + 1} \\ S_{n} & = & \sum_{1}^{n} (\frac{1}{n} - \frac{1}{n + 1}) \\ = & 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . . . . . + \frac{1}{n} - \frac{1}{n + 1} \\ = & \frac{n}{n + 1} \end{array}$