Question

CMR : \(1^2+2^2+3^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)\(\)

Answer 1

- Với \(n=1\Rightarrow1=\frac{1.2.3}{6}\) (đúng)

- Giả sử đẳng thức cũng đúng với \(n=k\) hay:

\(1^2+2^2+...+k^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}\)

Ta cần chứng minh nó cũng đúng với \(n=k+1\) hay:

\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)

Thật vậy, ta có:

\(1^2+2^2+...+k^2+\left(k+1\right)^2=\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)

\(=\left(k+1\right)\left(\frac{k\left(2k+1\right)}{6}+k+1\right)=\left(k+1\right)\left(\frac{2k^2+k+6k+6}{6}\right)\)

\(=\left(k+1\right)\left(\frac{2k^2+7k+6}{6}\right)=\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\) (đpcm)