The sum to n terms of the series113+1+213+23+1+2+313+23+33+…is

A
$\frac{n}{n + 1}$
B
$\frac{n}{2 (n + 1)}$
C
$\frac{2 n}{n + 1}$
D
$\frac{n}{n (n + 1)}$

Solution:

If $t_{n}$ denotes the nth term of the series, then

$\begin{array}{l} t_{n} = \frac{1 + 2 + 3 + \dots + n}{1^{3} + 2^{3} + 3^{3} + \dots + n^{3}} = \frac{\frac{1}{2} n (n + 1)}{\frac{1}{4} n^{2} (n + 1)^{2}} \\ = \frac{2}{n (n + 1)} = 2 [\frac{1}{n} - \frac{1}{n + 1}] \end{array}$

$\begin{array}{l} \Rightarrow \sum_{k = 1}^{n} t_{k} = 2 \{(\begin{array}{ll} 1 & \frac{1}{2} \end{array}) + (\begin{array}{cc} \frac{1}{2} & \frac{1}{3} \end{array}) + \dots + \\ (\frac{1}{n} - \frac{1}{n + 1})\} \\ = 2 (1 - \frac{1}{n + 1}) = \frac{2 n}{n + 1} \end{array}$

The sum to $n$ terms of the series
$\frac{1}{1^{3}} + \frac{1 + 2}{1^{3} + 2^{3}} + \frac{1 + 2 + 3}{1^{3} + 2^{3} + 3^{3}} + \dots$ is

Solution:

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