- Với \(n=1\) đúng
- Giả sử đúng với \(n=k\) hay: \(1^2+...+\left(2k-1\right)^2=\frac{k\left(4k^2-1\right)}{3}=\frac{k\left(2k-1\right)\left(2k+1\right)}{3}\)
Ta cần chứng minh nó đúng với \(n=k+1\) hay:
\(1^2+...+\left(2k-1\right)^2+\left(2k+1\right)^2=\frac{\left(k+1\right)\left[4\left(k+1\right)^2-1\right]}{3}=\frac{\left(k+1\right)\left(2k+1\right)\left(2k+3\right)}{3}\)
Thật vậy:
\(1^2+...+\left(2k-1\right)^2+\left(2k+1\right)^2=\frac{k\left(2k-1\right)\left(2k+1\right)}{3}+\left(2k+1\right)^2\)
\(=\left(2k+1\right)\left[\frac{k\left(2k-1\right)}{3}+2k+1\right]=\frac{\left(2k+1\right)\left(2k^2+5k+3\right)}{3}\)
\(=\frac{\left(2k+1\right)\left(k+1\right)\left(2k+3\right)}{3}=\frac{\left(k+1\right)\left(2k+1\right)\left(2k+3\right)}{3}\) (đpcm)
