Discussion Forum

 The sum of first 9 terms of the series $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...is$

Answer:

Explanation:

Central Idea. Write  the nth term of the given  series and simplify it to get its lowest forth. Then , apply , S_n = ∑ Tπ

Given series is 

  $\frac{1^{3}}{1}+\frac{1^{3}+2^{3}}{1+3}+\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+...\infty$

  Let T_n  be the nth term of the given series

   $\therefore$    $\frac{1^{3}+2^{3}+3^{3}+....+n^{3}}{1+3+5+.....+n terms}$

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

     = $\frac{(n+1)^{2}}{4}$

$S_{9}=\sum_{n=1}^{9}\frac{(n+1)^{2}}{4}=\frac{1}{4}[(2^{2}+3^{2}+.....+10^{2})$

                                                                     +  $1^{2}-1^{2}]$

  $=\frac{1}{4}\left[\frac{10(10+1)(20+1)}{6}-1\right]=\frac{384}{4}=96$