help faulhaber's formula summation problem I'm given that $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$
and told to prove $\sum_{k=1}^n (2k-1)=n^2$
$\sum_{k=1}^n (2k-1)=2\sum_{k=1}^n k- \sum_{k=1}^n 1=2\frac{n(n+1)}{2}-n$
$=2\frac{n(n+1)}{2}-n$ 
$=\frac{n\left(n+1\right)\cdot \:2}{2}-n$
$=n\left(n+1\right)-n$
$=n^2+n-n$
$=n^2$
$\sum_{k=1}^n (2k-1)=n^2$
next I need to write the summation that gives the sum
of the odd numbers found in the first $n$ rows of this sequence.
$1 = a_1$ 
$3 + 5 = a_2$ 
$7 + 9 + 11 = a_3$ 
$13 + 15 + 17 + 19 = a_4$ 
$21 + 23 + 25 + 27 + 29 = a_5$
$\vdots \vdots \vdots$
I'm givin that $n=3$ should yield the same result as $\sum_{k=1}^6 (2k-1)=36$
if you sub in $n^2$ to the formula for $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ you get $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}$
at $n=3$; $\frac{3^{2}\left(3+1\right)^{2}}{2^{2}}=36$
I'm having trouble getting that formula into the correct form I believe this is the end result but I'm unsure how to prove the equality $\frac{n^{2}\left(n+1\right)^{2}}{2^{2}}=\sum_{k=1}^{n}k^{3}$
Then I need to replace that sum with an explicit formula of n

Hint:  Write the rows of your sequence this way:
$$
\begin{aligned}
a_1 &= (2\cdot1 -1)\\
a_2 &= (2\cdot2-1) + (2\cdot3-1)\\
a_3 &= (2\cdot4-1) + (2\cdot5-1) + (2\cdot6-1)\\
a_4 &= (2\cdot7-1) + (2\cdot8-1) + (2\cdot9-1) + (2\cdot10-1)
\end{aligned}
$$
When you the sum the first $n$ rows of your sequence, you are summing $2k-1$ from $k=1$ up to a final value for $k$.
For $n=1$, the final value of $k$ is $1$.
For $n=2$, the final value of $k$ is $3$.
For $n=3$, the final value of $k$ is $6$.
For $n=4$, the final value of $k$ is $10$.
Find an expression for the final value of $k$, as a function of $n$. Then plug this final value into your formula
$$
\sum_{k=1}^{\text {final}}(2k-1) = ({\text {final}})^2.
$$
