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!
 A: 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.
A: It seems your issue is more conceptual than algebraic since you're stuck about which form of right hand side to use.
A proof by induction on a sum formula works by showing: (1) it holds in the base case, when the index is at its minimum; and (2) if it applies for the $n=k$ case, then it will also hold for the $n=k+1$ case.
With these two in hand we prove that the formula holds at any index. For example, we know it holds at $k=10$ because it holds at $k=1$ (via (1)) which implies it holds at $k=2$ (via (2)) which implies it holds at $k=3$ (via (2) again), and so on, repeatedly applying (2) until we reach 10.
We want to show that "the formula applies at $k$" implies that the formula holds at $k+1$, so our target is showing that $\sum_{i=1}^{k+1}i^3 = \frac{(k+1)^2((k+1)+1)^2}{4}$ and our ammunition is our assumption that $1^3+\ldots+k^3=\frac{k^2(k+1)^2}{4}$.
To connect the two, we notice that the left side of our target nests our assumption -- $\sum_{i=1}^{k+1}i^3 = (1^3+\ldots+k^3)+(k+1)^3$.
The rest is algebra.
A: 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.
A: 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.
