Make it simple

I wish to take reference to my question which is in a more general terms

Given that $n^2$ will appear $n$ times, the group of $n$ is essentially $n^3$ . The sum to it, however, will be valid only for the case the amount of terms satisfy $n = \frac{m(m+1)}{2}$ where m is a non-zero natural number, hence: $S_{\frac{n(n+1)}{2}}\ = (\frac{n(n+1)}{2})^2$ Provided that 99 did not satisfy this condition, we then will try to evaluate the closest approximation to it. Provided for $n = 13$ gives $S_{\frac{13(14)}{2}}\ = (\frac{13(14)}{2})^2$ or $S_{91}=8281$ . The next term will be $14^2 = 196$ which will continue from $a_{92}$ to $a_{105}$ . We only need up to $a_{99}$ , so the sum for the rest is $(99-92+1)\times14^2 = 1568$ . The sum for the first 99 terms of this series then be $8281+1568 = 9849$