Proof by mathematical induction: stuck on algebra

Question

Use mathematical induction to prove the following:

For all $i$ starting from $1$ up to $N$, $i^3 =\frac{N^2(N + 1)^2}{4}$

Base Case:

$1 ^ 3 = 1^2(1 + 1)^2 / 4$

Conditions holds, so move onto general case.

Assume true for $N = k$

$1 ^ 3 + 2 ^ 3 + 3 ^ 3 + ... + (k)^3 = k^2(k+1)^2 / 4$

Prove for $k + 1$

$k^2(k+1)^2 / 4 + (k + 1)^3 = k + 1 ^2(k+2)^2/4$

LHS:

$= k^2(k+1)^2 + (k+1)^3(4)/4$

$= k^4 + 2k^3 + k^2 + (k+1)^3(4) / 4$

$= k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4 / 4$

$= k^4 + 6k^3 + 13k^2 + 12k + 4 / 4$

So I get to this point, but I don't know how to proceed. I'm confident that what I've done so far is completely wrong.

It seems that you want to prove: $\sum_{i=1}^N i^3=\frac{N^2(N+1)^2}{4}$, correct? — Emilio Novati, May 31 '17 at 07:45
https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference mathjax reference. — Siong Thye Goh, May 31 '17 at 07:51

score 0 · Answer 1 · answered May 31 '17 at 07:43

0

We need to prove that $$\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$ or $$\frac{k^2}{4}+k+1=\frac{(k+2)^2}{4},$$ which is obvious.

answered May 31 '17 at 07:43

Michael Rozenberg

score 0 · Answer 2 · answered May 31 '17 at 07:45

0

If you can't figure out how to proceed from left hand side to right hand side via factorization.

Do consider working from right hand side as well.

Expand

\begin{align} (k+1)^2(k+2)^2&=(k^2+2k+1)(k^2+4k+4) \\ &=k^4+6k^3+13k^2+12k+4 \end{align}

answered May 31 '17 at 07:45

Siong Thye Goh

2 Answers2