Question

The sum of $n$ terms of the series $\frac{\frac{1}{2} \cdot \frac{2}{2}}{1^3}+\frac{\frac{2}{2} \cdot \frac{3}{2}}{1^3+2^3}+\frac{\frac{3}{2} \cdot \frac{4}{2}}{1^3+2^3+3^3}+\ldots \ldots .$. is $-$

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

Answer 1

Answer: (B)

$n ^{ th }$ term of the given series
$T _{ n }= \frac{\frac{ n }{2} \cdot \frac{ n +1}{2}}{\Sigma n ^3}=\frac{\frac{1}{4} n ( n +1)}{\frac{1}{4} n ^2( n +1)^2}=\frac{1}{ n ( n +1)}=\left(\frac{1}{ n }-\frac{1}{ n +1}\right) $
$\therefore S _{ n } =\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots+\left(\frac{1}{ n }-\frac{1}{ n +1}\right)\right]$
$=1-\frac{1}{ n +1}=\frac{ n }{ n +1}$