Use Mathematical Induction to prove equation? Use mathematical induction to prove each of the following statements.
let $$g(n) = 1^3 + 2^3 + 3^3 + ... + n^3$$
Show that the function $$g(n)= \frac{n^2(n+1)^2}{4}$$ for all n in N
so the base case is just g(1) right? so the answer for the base case is 1, because 4/4 = 1
then for g(2) is it replace all of the n's with n + 1 and see if there is a concrete answer?
 A: First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^3=\frac{1^2(1+1)^2}{4}$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^3=\frac{n^2(n+1)^2}{4}$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^3=$
$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$
$\color\red{\frac{n^2(n+1)^2}{4}}+(n+1)^3=$
$\frac{(n+1)^2(n+1+1)^2}{4}$

Please note that the assumption is used only in the part marked red.
A: Hint: Think of what you are supposed to prove: That a property $P(n)$ is true for every natural number. 
A proof by induction essentially proves that the property $P(n)$ holds for the first natural number and is preserved, as we count upwards.
In this case the property $P(n)$ is 
$$P(n) \text{ is true if } g(n) = \frac{n^2(n+1)^2}{4}$$
The base case consists of proving that the property $P(1)$ is true, that is, that
$$g(1) = \frac{1^2(1+1)^2}{4}$$
In the inductive step you must prove that if $P(n)$ is true, then so is $P(n+1)$.
Now what does it mean that $P(n+1)$ is true?
A: You want to show that $$\sum_{j=1}^n j^3 = \frac{n^2(n+1)^2}{4}.$$
You've shown that it holds for the base case, which is good.
Now, you assume that it holds for $n=k$:
$$\sum_{j=1}^k j^3 = \frac{k^2(k+1)^2}{4}.$$
Then, a strategy is to write out one part or the other for the $n=k+1$ case, and manipulate it so that you pull out the $n=k$ expression.  Let's do the sum:
$$\sum_{j=1}^{k+1} j^3 = \sum_{j=1}^{k} j^3 + (k+1)^3.$$
This is just algebra; I pulled out the last term and wrote it explicitly.  But now I can substitute in the assumption for the sum on the right hand side.
The rest is to manipulate the expression on the right hand side to come up with the expression
$$\frac{(k+1)^2(k+2)^2}{4},$$
and then you're done.
Can you take it from here?
A: Show that: $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$

$(1)$ First step is to evaluate the expression for $n=1$, $$1^3=\frac{1\times(1+1)^2}{4}$$
Which is $1=1$, which means we can proceed.

$(2)$ Now assume it is true for $n$, $$1^3 + 2^3 + 3^3 + ... + n^3= \frac{n^2(n+1)^2}{4}$$

$(3)$ Lastly prove for $n+1$ and you have proved it for all $n\in\mathbb N$,
$$1^3 + 2^3 + 3^3 + ... +n^3+ (n+1)^3= \frac{(n+1)^2(n+2)^2}{4}$$
After "inserting" $n+1$, replace the left side as it follows:
$$ \frac{n^2(n+1)^2}{4}+(n+1)^3=\frac{(n+1)^2(n+2)^2}{4}$$
Simply "tidy up" the left side to get the final form;
$$\frac{(n+1)^2(n+2)^2}{4}=\frac{(n+1)^2(n+2)^2}{4}$$

And that's an example of a proof with induction.
A: Base Case:
Show that $g(n)$ holds for $n=1$.
(You've done this already).

Inductive Hypothesis:
Assume that $g(n)$ holds for arbitrary $k \in \mathbb{N}$, where $k \geq 1$ (since you already proved base case).
Thus we have that:
$$g(k) = \frac{k^2(k+1)^2}{4}$$

Inductive Step:
Show that $g(k) \to g(k+1)$
$$g(k+1) = \frac{(k+1)^2((k+1)+1)^2}{4} = \frac{(k+1)^2(k+2)^2}{4}$$
Notice that if we expand $g(k+1)$, we get:
$$\frac{\left(k^4+6k^3+13k^2+12k+4\right)}{4}$$
Now for the proof:
We know that $g(k+1) = g(k) + (k+1)^3$
We can assume $g(k)$ holds via our inductive hypothesis.
Thus we have:
$$g(k+1) = g(k) + (k+1)^3$$
$$g(k+1) = \frac{k^2(k+1)^2}{4} + (k+1)^3$$
$$g(k+1) = \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4}$$
After you simplify and collect like terms, you will end up with $g(k+1)$ (this is why I expanded the original equation for $g(k+1)$ so we don't have to factor anything hard.
QED.
A: Assuming you are asking what the inductive step is, you simply need to assume that $g(n) =\frac{n^2(n+1)^2}{4}$ and then use it to evaluate g(n+1) like so:
\begin{equation}
g(n+1) = g(n) + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{n^2(n+1)^2}{4} + (n+1)^3
\end{equation}
\begin{equation}
g(n+1) = \frac{4(n+1)^3 + n^2(n+1)^2}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(4(n+1) + n^2)}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(n^2 + 4n + 4)}{4}
\end{equation}
\begin{equation}
g(n+1) = \frac{(n+1)^2(n+2)^2}{4}
\end{equation}
Q.E.D.
