Đặt \(A=1\cdot2+2\cdot3+\cdots+99\cdot100\)
\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)
\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)
\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\left(99+1\right)}{2}\)
\(=\frac{99\cdot100\cdot199}{6}+\frac{3\cdot99\cdot100}{6}=\frac{99\cdot100\cdot202}{6}\)
\(=33\cdot50\cdot202=333300\)
\(x+\left(x+1\cdot2\right)+\left(x+2\cdot3\right)+\cdots+\left(x+99\cdot100\right)=0\)
=>x+99x+333300=0
=>100x=-333300
=>x=-3333
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