Let's set $S_{n}=1^{3}+\cdots+n^{3}$ and $u_{n}=1+3+5+\cdots+(2 n-1)$ the sum of the first $n$ odd numbers.

1. By evaluating for small $n$, predict and then prove by induction a formula for $u_{n}$.
2. Using the formula found above for the sum of linear terms, find $u_{n}$ again.
3. Can we find a third way to derive a formula for $u_{n}$?
4. By noticing that $1^{3}=1, 2^{3}=3+5, 3^{3}=7+9+11$ etc., find a formula for $S_{n}$.
1. Notice that $u_{1}=1, u_{2}=4, u_{3}=9$ and $u_{4}=16$. It seems legitimate to think that $u_{n}=n^{2}$. Let's prove this by induction.

Initialization: For $n=1, u_{1}=1^{2}$

Hereditary: Let $n \in \mathbb{N}^{*}$ such that $u_{n}=n^{2}$. Then

$$
u_{n+1}=u_{n}+(2(n+1)-1)=u_{n}+2 n+1 \underset{\text { H.R. }}{=} n^{2}+2 n+1=(n+1)^{2}
$$

Hence the result by induction.

2. Notice that

$$
\begin{gathered}
u_{n}=1+3+\cdots+(2 n-1)=(2 \cdot 1-1)+(2 \cdot 2-1)+\cdots+(2 n-1)=2 \cdot(1+\cdots+n)-\underbrace{(1+\cdots+1)}_{n \text { times }} \\
=2 \cdot(1+\cdots+n)-n
\end{gathered}
$$

The example from the course provides a formula for the first sum, let's use it:

$$
u_{n}=2 \cdot(1+\cdots+n)-n=2 \cdot \frac{n(n+1)}{2}-n=n(n+1)-n=n^{2}
$$

3. Consider the following drawing:

We notice that there is 1 red point, 3 orange points, 5 blue points, etc., up to the $n$-th odd number. Similarly, it is simply an $n \times n$ square. We thus obtain that $u_{n}=n^{2}$.

4. The remark gives us that $S_{n}$ is the sum of the odds associated with $1^{3}$, then with $2^{3}$ up to $n^{3}$. We notice that the odds are successive and do not repeat, so we end up with $S_{n}$ being the sum of all the odds from 1 to the last added by $n^{3}$, i.e., that $S_{n}=u_{k}$ with $k$ the total number of odd terms.

Let's calculate $k$:

$1^{3}$ adds one odd number, $2^{3}$ adds 2 odd numbers, $3^{3}$ adds 3 odd numbers, etc. We thus have, in total, $k=1+2+\cdots+n$ odd terms.

Thus $S_{n}=u_{k}=k^{2}=(1+\cdots+n)^{2}=\left(\frac{n(n+1)}{2}\right)^{2}$