Question

Tính m = 1² + 2² + 3² +.....+ 99² + 100²

Answer 1

Ta có:

\(M=1^2+2^2+3^2+...+99^2+100^2\)

\(=1+2\left(1+1\right)+...+99\left(98+1\right)+100\left(99+1\right)\)

\(=1+1.2+2+...+98.99+99+99.100+\) \(100\)

\(=\left(1.2+2.3+...+99.100\right)\) \(+\left(1+2+...+99+100\right)\)

\(=333300+5050=338050\)

\(\Rightarrow\) Tổng quát: \(M=\dfrac{\left(n-1\right)n\left(n+1\right)}{3}+\dfrac{n\left(n+1\right)}{2}\)

Answer 2

Áp dụng \(1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\) ta có:

\(M=1^2+2^2+3^2+...+100^2\)

\(=\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}=338350\)