Question

Tính A=1^2+2^2+3^2+4^2+...+100^2

Giải giúp mk nhanh nhé các bạn ! Thank you very much.

Answer 1

\(A=1+2\left(1+1\right)+3\left(2+1\right)+4\left(3+1\right)+...+100.\left(99+1\right).\)

\(A=1+1.2+2+2.3+3+3.4+4+...+99.100+100\)

\(A=\left(1+2+3+4+...+100\right)+\left(1.2+2.3+3.4+...+99.100\right)\)

\(B=1+2+3+4+...+100=\frac{100\left(1+100\right)}{2}=5050\)

\(C=1.2+2.3+3.4+...+99.100\)

\(3C=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3C=1.2.3+2.3\left(4-1\right)+3.4\left(5-2\right)+...+99.100\left(101-98\right)\)

\(3C=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3C=99.100.101\Rightarrow C=\frac{99.100.101}{3}=33.100.101=333300\)

\(A=B+C=5050+333300=338350\)