Prove the following by using the principle of mathematical induction for all `n in N`:`1^3+2^3+3^3+...........+n^3=((n(n+1))/2)^2`

Explanation

`P(n) : 1^(3) + 2^(3) + 3^(3) + "…. + n^(3) = (n^(2)(n+1)^(2))/(4)`
For `n = 1`,
L.H.S. `= 1^(3) = 1` and `R.H.S. = (1^(2).2^(2))/(4) = 1`
Thus, `P(1)` is true.
Let `P(n)` be true for some `n = k`.
i.e, `1^(3) + 2^(3) + 3^(3) + "...." + k^(3) = (k^(2) (k+1)^(2))/(4) "....."(1)`
Now, we have to prove that ` P(n)` is true for ` n = k + 1`.
i.e, `1^(3) + 2^(3) + 3^(3) + ".... " +k^(3) + (k+1)^(3) = ((k+1)^(2)(k+2)^(2))/(4)`
Adding `(k+1)^(3)` on both sides of `(1)`, we get `1^(3) + 2^(3) + 3^(3) + ".... " +k^(3) + (k+1)^(3)`
`= (k^(2)(k+1)^(2))/(4)+(k+1)^(3)`
`= ((k+1)^(2))/(4) (k^(2)+4(k+1))`
`= ((k+1)^(2)(k+2)^(2))/(4)`
Thus, `P (k+1)` is true whenever `P(k)` is true.
Hence, by the principle of malthematical induction, statement `P(n)` is true for all natural numbers.