summation - 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!

Answer

Your inductive assumption is such that the formula marked $\color{red}{\mathrm{red}}$ (several lines below) holds for $i=k$: $$\sum^{i=k}_{i=1} i^3=\frac{k^2 (k+1)^2}{4}$$

You need to prove that for $i=k+1$: $$\sum^{i=k+1}_{i=1} i^3=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$
To do this you cannot use: $$\sum^{i=n}_{i=1} i^3=\color{red}{\frac{n^2 (n+1)^2}{4}}$$ as this is what you are trying to prove.

So what you do instead is notice that:
$$\sum^{i=k+1}_{i=1} i^3= \underbrace{\frac{k^2 (k+1)^2}{4}}_{\text{sum of k terms}} + \underbrace{(k+1)^3}_{\text{(k+1)th term}}$$
$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{1}{4}k^2+(k+1)\right)$$

$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{k^2+4k+4}{4}\right)$$
$$\sum^{i=k+1}_{i=1} i^3= (k+1)^2\left(\frac{(k+2)^2}{4}\right)=\color{blue}{\frac{(k+1)^2 (k+2)^2}{4}}$$

Which is the relation we set out to prove. So the method is to substitute $i=k+1$ into the formula you are trying to prove and then use the inductive assumption to recover the $\color{blue}{\mathrm{blue}}$ equation at the end.