\(B=1\cdot2^2+2\cdot3^2+3\cdot4^2+...+99\cdot100^2\\ =\left(2-1\right)\cdot2^2+\left(3-1\right)\cdot3^2+\left(4-1\right)\cdot4^2+...+\left(100-1\right)\cdot100^2\\ =2\cdot2^2-1\cdot2^2+3\cdot3^2-1\cdot3^2+4\cdot4^2-1\cdot4^2+...+100\cdot100^2-1\cdot100^2\\ =2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\ =1^3-1^2+2^3-2^2+3^3-3^2+4^3-4^2+...+100^3-100^2\\=\left(1^3+2^3+3^3+4^3+...+100^3\right)-\left(1^2+2^2+3^2+4^2+...+100^2\right)\\=\left(1+2+3+...+100\right)^2-\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\\ =\left[\dfrac{\left(100\cdot101\right)}{2}\right]^2-\dfrac{100\cdot101\cdot201}{6}\\ =5050^2-2030100\\ =25502500-2030100\\ =23472400 2\)
Bonus: Công thức:
1) \(1+2+3+...+n=\dfrac{n\cdot\left(n+1\right)}{2}\)
2) \(1^2+2^2+3^2+...+n^2=\dfrac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}\)
3) \(1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)
