\(1^2+2^2+3^2+...+99^2\\ =1\cdot\left(0+1\right)+2\cdot\left(1+1\right)+3\cdot\left(2+1\right)+...+99\cdot\left(98+1\right)\\ =1+1\cdot2+2+2\cdot3+3+...+98\cdot99+99\\ =\left(1\cdot2+2\cdot3+...+98\cdot99\right)+\left(1+2+3+...+99\right)\\ 1+2+3+...+99=\dfrac{99\cdot100}{2}=4950\\ 1\cdot2+2\cdot3+...+98\cdot99\\ =\dfrac{3\left(1\cdot2+2\cdot3+...+98\cdot99\right)}{3}\\ =\dfrac{1\cdot2\cdot3+2\cdot3\cdot3+...+98\cdot99\cdot3}{3}\\ =\dfrac{1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+...+98\cdot99\cdot\left(100-97\right)}{3}\\ =\dfrac{1\cdot2\cdot3-0+2\cdot3\cdot4-1\cdot2\cdot3+...+98\cdot99\cdot100-97\cdot98\cdot99}{3}\\ =\dfrac{98\cdot99\cdot100}{3}\\ =323400\\ 1^2+2^2+3^2+...+99^2\\=4950+323400\\=328350\)
\(a=1^2+2^2+3^2+4^2+...+100^2\)
\(a=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+100.\left(101-1\right)\)
\(a=1.2-1+2.3-2+3.4-3+...+100.101-100\)
\(a=\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)\)
Đặt: \(\left\{{}\begin{matrix}l=1.2+2.3+3.4+...+100.101\\n=1+2+3+...+100\end{matrix}\right.\)
Áp dụng tính ta được: \(l=1.2+2.3+3.4+...+100.101\)
\(3l=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+100.101.\left(102-99\right)\)
\(3l=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+100.101.102-99.100.101\)
\(3l=100.101.102\Leftrightarrow l=\dfrac{100.101.102}{3}=343400\)
\(n=1+2+3+...+100\)
\(n=\left[\left(100-1\right):1+1\right]:2.\left(100+1\right)=50.101=5050\)
\(a=l-n=343400-5050=338350\)
