\(B=1+\dfrac{1}{2}\left(1+2\right)+...+\dfrac{1}{x}\left(1+2+..+x\right)\)
\(B_x=\dfrac{1}{x}\left(\dfrac{x\left(x+1\right)}{2}\right)=\dfrac{x+1}{2}\)
\(2B=2+3+4+5+...+\left(x+1\right)\)
\(2B+1=1+2+...+\left(x+1\right)=\dfrac{\left(x+1\right)\left(x+2\right)}{2}\)
\(B=115\Leftrightarrow2B+1=231\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=231.2=462\)=21.22
x=20