$B=1+\genfrac{}{}{0ex}{}{1}{2}\left(1+2\right)+\genfrac{}{}{0ex}{}{1}{3}\left(1+2+3\right)+\genfrac{}{}{0ex}{}{1}{4}\left(1+2+3+\text{4}\right)+...+\genfrac{}{}{0ex}{}{1}{x}\left(1+2+3+...+x\right)$

$B=1+\genfrac{}{}{0ex}{}{1}{2}\left(\genfrac{}{}{0ex}{}{2.3}{2}\right)+\genfrac{}{}{0ex}{}{1}{3}\left(\genfrac{}{}{0ex}{}{3.4}{2}\right)+\genfrac{}{}{0ex}{}{1}{4}\left(\genfrac{}{}{0ex}{}{4.5}{2}\right)+...+\genfrac{}{}{0ex}{}{1}{x}\left(\genfrac{}{}{0ex}{}{x\left(x+1\right)}{2}\right)$

$B=1+\genfrac{}{}{0ex}{}{3}{2}+\genfrac{}{}{0ex}{}{4}{2}+\genfrac{}{}{0ex}{}{5}{2}+...+\genfrac{}{}{0ex}{}{x+1}{2}$

$B=\genfrac{}{}{0ex}{}{1}{2}\left(2+3+4+...+\left(x+1\right)\right)$

$B=\genfrac{}{}{0ex}{}{1}{2}\left(\genfrac{}{}{0ex}{}{x\left(x+3\right)}{2}\right)$

Từ đó B=115 khi $\genfrac{}{}{0ex}{}{1}{2}\left(\genfrac{}{}{0ex}{}{x\left(x+3\right)}{2}\right)$=115 $\iff$$x\left(x+3\right)=460$

theo công thức tính tổng