Prove that :`1^2+2^2+3^2++n^3={(n(n+1))/2}^2dot`

Explanation

Let `P(n):1^3+2^3+3^3+.....+n^3=[(n(n+1))/(2)]^2`......(i)
Step I For `n=1`, LHS of Eq.(i) `=^3=1` and RHS of Eq. (i). `[(1(1+1))/(2)]^2=1^2=1`
`therefore LHS=RHS`
Therefore ,P(1), is ture.
Step II Assume P(k) is true , then
`P(k):1^3+2^3+3^3+.....K^3=[(k(k+1))/(2)]^2`
Step III For `n=k+1`,
`P(k+1):1^3+2^3+3^3+......+^3(k+1)^3`
`=[((k+1)+(k+2))/(2)] ^2`
LHS `=1^3+2^3+3^3+.....+k^3+(k+1)^3=[(k(k+1))/(2)]^2+(k+1)^3` [by assumption step]
`=((k+1)^2)/(4)[k^2+4(k+1)]`
`=((k+1)^2(k^2+4k+4))/(4)`
`=((k+1)^2(k+2)^2)/(4)`
`=[(k+1(k+2))/(2)]^2=RHS`
Therefore , `P(k+1)` is true , Hence , by the principle of mathematical induction , P(n)is true for all `n epsi N`.