\(C=1\cdot99+2\cdot98+3\cdot97+...+98\cdot2+99\cdot1\)
\(C=\left(1+2+3+...+98+99\right)\left(99+98+...+3+2+1\right)\)
Mà \(\left(1+2+3+...+98+99\right)=\left(99+98+...+3+2+1\right)\)
\(\Rightarrow C=\left(1+2+3+...+98+99\right)^2\)
Tính \(1+2+3+...+98+99\)
\(=\left(99+1\right)+\left(98+2\right)+\left(97+3\right)+.....\)
\(=100\cdot\frac{99}{2}=4950\)
Có \(C=\left(1+2+3+...+98+99\right)^2\)
\(\Rightarrow C=4950^2\)