<s>[INST] 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!
 [/INST] You want to prove that $$\sum_{i=1}^{n}i^3 = \frac{n^2 (n+1)^2}{4}$$ using induction.
For $n=1$, $$\sum_{i=1}^{1}i^3 = 1^3=1=\frac{1^2(1+1)^2}4=1$$
So the formula does work for the base case $n=1$.
Now, assume the formula works for $n=k$ and show that this implies that the formula is correct for $n=k+1$ which will accomplish the prove by induction.
Thus, 
\begin{align}
\sum_{i=1}^{k+1}i^3 & = \left(\sum_{i=1}^{k}i^3\right) +(k+1)^3 \\
& = \frac{k^2 (k+1)^2}{4}+(k+1)^3 \\
& = \frac{k^2 (k+1)^2+4(k+1)^3}{4} \\
& = \frac{ (k+1)^2(k^2+4(k+1))}{4} \\
& = \frac{ (k+1)^2(k+2)^2}{4} \\
& = \frac{ (k+1)^2((k+1)+1)^2}{4}
\end{align}
and you are done.
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