Question

Show that $1^2+2^2+3^2+\dots \dots \dots \dots \dots +n^2=\frac{n(n+1)(2n+1)}{6}$ for all $n\in N$

Answer 1

$check{n}^3-(n-1{)}^3=n^3-[n^3-3n^2+3n-1]=3n^2-3n+1\therefore 1^3-0^3=3\times 1^2-3\times 1+12^3-1^3=3\times 2^2-3\times 2+13^3-2^3=3\times 3^2-3\times 3+1.........n^3-(n-1{)}^3=3\times n^2-3\times n+1n^3=3(1^2+2^2+....+n^2)-3(1+2+...+n)+nor{n}^3+3(1+2+.....+n)-n=3(1^2+2^2+....+n^2)n^3+3\left(\frac{n(n+1)}{2}\right)-n=3(1^2+2^2+....+n^2)n\left[\right(\frac{2n^2+3n+3-2}{2}\left)\right]=3(1^2+2^2+....+n^2)\left(\frac{n(n+1)(2n+1)}{2}\right)=3(1^2+2^2+...+n^2)\therefore 1^2+2^2+.....+n^2=\left(\frac{n(n+1)(2n+1)}{6}\right)$