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`

Prove the following by using the principle of mathematical induction f

1.	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`
Answer» Let the given statement be P(n). Then, ` P(n) : 1^(3)+2^(3)+3^(3)+…+n^(3)= {(n(n+1))/2}^(2)`. Putting n = 1 in the given statement, we get LHS = `1^(3) = 1 and RHS = ((1xx2)/2)^(2) = 1^(2) = 1`. `:. ` LHS = RHS . Thus, P(1) is true. Let P(k) be for some `k in N`. Then, `P(k) : 1^(3)+2^(3)+3^(3) + ... k^(3) = {(k(k+1)^(2))/2}^(2)." "` ...(i) Now, `1^(3)+2^(3)+3^(3)+...+k^(3)+(k+1)^(3)` ` = {1^(3)+2^(3)+3^(3)+...+k^(3)}+(k+1)^(3)` ` = {(k(k+1))/2}^(2) +(k+1)^(3)" "` [using (i)] `=(k+1)^(2){k^(2)/4+(k+1)}=(k+1)^(2){(k^(2)+4k+4)/4}` `=((k+1)^(2)(k+2)^(2))/4 = {((k+1){(k+1)+1})/2}^(2)` ` :. " " P(k+1): 1^(3)+2^(3)+3^(3)+...+(k+1)^(3)=[((k+1){(k+1)+1})/2]^(2)`. This shows that P(k+1) is true, whenever P(k) is true. `:." " P(1) ` is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, we have ` 1^(3)+2^(3)+3^(3)+...+n^(3)={(n(n+1))/2}^(2)" for all " n in N`.