If `S_1,S_2,S_3`are the sum of first n natural numbers, their squares and their cubes, respectively, show that `9S_2^2=S_3(1+8S_1)`.

Explanation

It is given that
`S_1` is the sum of `n` natural numbers

i.e. `S_1=1+2+3+......+n`

`S_1=(n(n+1))/2`


`S_2` is the sum of square of `n` natural numbers

i.e. `S_2=1^2+2^2+3^2+......+n^2`

`S_2=(n(n+1)(2n+1))/6`


Also `S_3` is the sum of their cubes

i.e. `S_3=1^3+2^3+3^3+......+n^3`

`S_3=((n(n+1))/2)^2`

`S_3=(n^2(n+1)^2)/4`


We need to Show that `9S_2^2=S_3(1+8S_1)`

Taking R.H.S.
`S_3(1+8S_1)`

` " " " " = (n^2(n+1)^2)/4(1+8((n(n+1))/2))`

` " " " " = (n^2(n+1)^2)/4(1+4n(n+1))`

` " " " " = (n^2(n+1)^2)/4(1+4n^2+4n)`

` " " " " = (n^2(n+1)^2)/4((2n)^2+(1)^2+2xx2nxx1)`

` " " " " = (n^2(n+1)^2)/4(2n+1)^2`

` " " " " = ((n(n+1)(2n+1))^2)/4`

Taking L.H.S.
`9S_2^2`

` " " " " = 9((n(n+1)(2n+1))/6)^2`

` " " " " = 9((n(n+1)(2n+1))^2/6^2)`

` " " " " = 9xx(n(n+1)(2n+1))^2/36`

` " " " " = (n(n+1)(2n+1))^2/4`

` " " " " =` R.H.S.

L.H.S.= R.H.S.

Hence Proved