If `S_1,S_2,S_3` are the sums of first n natural numbers, their squares and their cubes respectively then `S_3(1+8S_1)=`

Answer

N/a

Explanation

`S_(1) = sumn = (1)/(2)n(n + 1)`
`S_(2) = sumn^(2) = (1)/(6)n(n + 1)(2n + 1)`
`S_(3) = sumn^(3) = (1)/(4)n^(2)(n + 1)^(2)`
R.H.S. = `S_(3)(1 + 8S_(1))`
`= (1)/(4)n^(2)(n + 1)^(2) * [1 + 8 * (1)/(2)n(n + 1)]`
`= (1)/(4)n^(2)(n + 1)^(2)(1 + 4n^(2) + 4n)`
`= (1)/(4)n^(2)(n + 1)^(2)(2n + 1)^(2)`
`= 9 * (1)/(36)n^(2)(n + 1)^(2)(2n + 1)^(2)`
`= 9 * [(1)/(6)n(n + 1)(2n + 1)]^(2)`
`9 * S_(2)^(2) = L.H.S.` Hence Proved.