Alg8.9

Find the sum of the squares of the first $n\,$ natural numbers $1,2,3,4.........\,$

Observe the following identity

$x(x+1)(2x+1)-(x-1)x(2x-1)=6x^2\,$

If $x=1\,$ then $1.2.3-0.1.1=6.1^2\,$

If $x=2\,$ then $2.3.5-1.2.3=6.2^2\,$

If $x=3\,$ then $3.4.7-2.3.5=6.3^2\,$ etc.

If $x=(n-1)\,$ then $(n-1)n(2n-1)-(n-2)(n-1)(2n-3)=6\cdot(n-1)^2\,$

If $x=n\,$ then $n(n+1)(2n+1)-(n-1)n(2n-1)=6\cdot n^2\,$

Adding all the terms, we get $n(n+1)(2n+1)=6[1^2+2^2+3^2+.......+n^2]=6\sum n^2\,$

Therefore $\sum n^2=\frac{n(n+1)(2n+1)}{6}\,$