Question

Tính tổng :\(1^2+2^2+3^2+...+100^2\)

Answer 1

Tham Khao

Answer 2

\(1^2+2^2+3^2+...+100^2=1.\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)=1.2-1+2.3-2+3.4-4+...+100.101-100=\left(1.2+2.3+3.4+...+100.101\right)-\left(1+2+3+...+100\right)=\dfrac{3\left(1.2+2.3+3.4+...+3.100.101\right)}{3}-\left(1+2+3+...+100\right)=\dfrac{1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+100.101\left(102-99\right)}{3}-\dfrac{\left(100+1\right)\left(\dfrac{100-1}{1}+1\right)}{2}=\dfrac{1.2.3-1.2.3+2.3.4-2.3.4+...+-99.100.101+100.101.102}{3}-5050=\dfrac{100.101.102}{3}-5050=343400-5050=338350\)