Write on my own my first mathematical induction proof I am trying to understand how to write mathematical induction proofs. This is my first attempt.
Prove that the sum of cubic positive integers is equal to the formula 
$$\frac{n^2 (n+1)^2}{4}.$$ I think this means that the sum of cubic positive integers is equal to an odd number. However, let's go on proving...
1) I start by proving the base case $n=1$ and I show that the formula holds.
2) I assume than any number $k$ other than $1$, which appartains at $N$, holds for the formula and I write the same formula but with $k$ which replaces $n$.
3) For mathematical induction, I assume that the formula holds also for $k+1$ = $n$ 
So, the left side of the equation should be:
$$\sum^{k+1}_{i=1} i^3 = 1^3 + 2^3 + 3^3 + ... + (k+1)^3$$
I am wondering about which one of these 2 forms (equivalents, I think) should have the right side :
this one, with $k+1$ in place of the $n$ of the original formula / or $k$ in the second version:  $\frac{(k+1)^2[(k+1)+1]^2}{4}$  or this one: $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ ?
I think that, in order for the proof to be convincing, we should write an equivalent statement for the original form of the formula, namely $$\sum^{n}_{i=1} i^3= \frac{n^2(n+1)^2}{4}$$ and perhaps we do it by showing that after algebraic passages $\frac{k^2(k+1)^2 }{4} + (k+1)^3$ is equal to $\frac{(k+1)^2[(k+1)+1]^2}{4}$ ?
Sorry for my soliloquy but it helps to understand and I would appreciate confirmation from you!

You should use the second one:
Suppose it holds for the first $k$ numbers. So their sum is equal to $\frac{k²(k+1)^2}{4}$. Then the first sum of the first $k+1$ is equal to $1^3 + 2^3 + 3^3 + ... + (k+1)^3=\frac{k²(k+1)^2}{4}+(k+1)^3=\frac{k²(k+1)^2}{4}+\frac{4(k+1)^3}{4}$ which is equal to $\frac{k²(k+1)^2+4(k+1)^3}{4}=\frac{(k+1)²(k²+4k+4)}{4}=\frac{(k+1)²(k+2)²}{4}$.
Which is precisely what you need.
