JEE-Mains 2015 (9/30)

The $n^{th}$ term of this sequence can be represented as $\frac{\sum_{i=1}^n i^3}{1+3+5... + (2n-1)}$ Sum of odd natural numbers is $n^2$ .

Let $x_i$ denote the $i^{th}$ term of the given equation : $\sum x_i = \sum_{i=1}^n\frac{(\frac{(i)(i+1)}{2})^2}{i^2}$ $\sum x_i = \sum_{i=1}^n\frac{(i+1)^2}{4}$ Expanding $(n+1)^2$ : $\sum_{i=1}^{n} x_i = \frac{\sum_{i=1}^n i^2 + 2 \times \sum_{i=1}^n i + n}{4}$ $\sum_{i=1}^{n} x_i = \frac{1}{4}(\frac{n\times(n+1)\times(2n+1)}{6} + 2 \times \frac{n\times(n+1)}{2} + n)$ To find the sum of the first 9 terms, we simply plug in $9$ to get $\sum_{i=1}^{9} x_i = \frac{1}{4}(285 + 90 + 9 )$ $\large\boxed{\sum_{i=1}^{9} x_i = 96}$

$n^{th} \space term = \frac{\sum n^3}{n^2}$

$n^{th} \space term = \frac{\left[ \frac{n(n+1)}{2}\right] ^2}{n^2}$

$\sum n^{th} \space term = \sum \frac{(n+1)^2}{4}$

$\sum n^{th} \space term =\frac{1}{4}\left( \sum n^2 + 2\sum n + \sum 1\right)$

$\Rightarrow \frac{1}{4} \left( \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} +n\right)$

$\text{put n = 9}$ $\boxed{96}$