KVPY #7

Each term $a_n$ is equal to:

$\large \displaystyle a_n = \frac{1^3 + 2^3 + .. n^3}{1 + 3 + ... 2n-1}$

$\large \displaystyle a_n = \frac{\Big (\frac{ n(n+1)}{2} \Big) ^2 }{ \frac{(2n-1 + 1)n}{2} }$

$\color{#20A900} \boxed { \large \displaystyle a_n = \frac{ (n+1)^2}{4} }$

The sum of the first $K$ terms is:

$\large \displaystyle S_K = \sum_{n=1}^K a_n$

$\large \displaystyle S_K = \frac14 \sum_{n=1}^K (n+1)^2$

Setting $m = n+1$

$\large \displaystyle S_K = \frac14 \sum_{m=2}^{K+1} m^2$

$\large \displaystyle S_K = \frac14 \left( \sum_{m=1}^{K+1} m^2 - 1 \right )$

$\large \displaystyle S_K = \frac14 \left [ \frac{(K+1)(K+2)(2K+3)}{6} - 1 \right ]$

$\color{#20A900} \boxed { \large \displaystyle S_K = \frac{2K^3 + 9K^2 + 13K}{24} }$

We're looking for $S_9$ :

$\large \displaystyle S_9 = \frac{2 \cdot 9^3 + 9 \cdot 9^2 + 13 \cdot 9}{24}$

$\color{#3D99F6} \boxed { \large \displaystyle S_9 = 96}$

$\text{1st term}:$ $\frac{1^2}{1^2} = 1$

$\text{2nd term}:$ $\frac{3^2}{2^2} = \frac{9}{4}$

$\text{3rd term}:$ $\frac{6^2}{3^2} = \frac{36}{9}$

$\text{4th term}:$ $\frac{10^2}{4^2} = \frac{100}{16}$

$\text{5th term}:$ $\frac{15^2}{5^2} = \frac{225}{25}$

$\text{6th term}:$ $\frac{21^2}{6^2} = \frac{441}{36}$

$\text{7th term}:$ $\frac{28^2}{7^2} = \frac{784}{49}$

$\text{8th term}:$ $\frac{36^2}{8^2} = \frac{1296}{64}$

$\text{9th term}:$ $\frac{45^2}{9^2} = \frac{2025}{81}$

$sum = 1 + \frac{9}{4} + \frac{36}{9} + \frac{100}{16} + \frac{225}{25} + \frac{441}{36} + \frac{784}{49} + \frac{1296}{64} + \frac{2025}{81} =$ $\color{#624F41}\large\boxed{96}$