Alg8.10

Find the sum of the cubes of the first $n\,$ natural numbers.

Observe the following identity

$x^2(x+1)^2-(x-1)^2 x^2\equiv 4x^3\,$

If $x=1\,$ then $1^2\cdot 2^2-0^2\cdot1^2=4\cdot1^3\,$

If $x=2\,$ then $2^2\cdot3^2-1^2\cdot2^2=4.2^3\,$

If $x=3\,$ then $3^2\cdot4^2-2^2\cdot3^2=4\cdot3^3\,$ etc.

If $x=n-1\,$ then $(n-1)^2 n^2-(n-2)^2(n-1)^2=4(n-1)^3\,$

If $x=n\,$ then $n^2(n+1)^2-(n-1)^2 n^2=4n^3\,$

Adding the terms we get $n^2(n+1)^2=4[1^3+2^3+3^3+.....+n^3]=4\sum n^3\,$

Therefore $\sum n^3=\frac{n^2(n+1)^2}{4}=[\frac{n(n+1)}{2}]^2\,$ or $[\sum n]^2\,$