Ta có: \(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\left(1\right)\)
Ta đi chứng minh:
*)Với \(n=1\) thì \(\left(1\right)\) đúng
Giả sử \(\left(1\right)\) đúng với \(n=k\), khi đó \(\left(1\right)\) thành
\(1^2+2^2+3^2+...+k^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Thật vậy giả sử \(\left(1\right)\) đúng với \(n=k+1\) khi đó \(\left(1\right)\) thành
\(1^2+2^2+...+k^2+\left(k+1\right)^2=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\left(2\right)\)
Cần chứng minh \(\left(2\right)\) đúng:
\(1^2+2^2+...+k^2+\left(k+1\right)^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(\Rightarrow1^2+2^2+...+k^2+\left(k+1\right)^2=\dfrac{k\left(k+1\right)\left(2k+1\right)}{6}+\dfrac{6\left(k+1\right)^2}{6}\)
\(=\dfrac{\left(k+1\right)\left[k\left(2k+1\right)+6\left(k+1\right)\right]}{6}=\dfrac{\left(k+1\right)\left[2k^2+k+6k+6\right]}{6}\)
\(=\dfrac{\left(k+1\right)\left[\left(2k^2+3k\right)+\left(4k+6\right)\right]}{6}=\dfrac{\left(k+1\right)\left[k\left(2k+3\right)+2\left(2k+3\right)\right]}{6}\)
\(=\dfrac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\). Suy ra \(\left(2\right)\). Theo nguyên lí quy nạp ta có ĐPCM
