<s>[INST] 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
 [/INST] Note that $a_i$ is a sum of $i$ terms. The first term is $(i^2 - i + 1)$ and the common difference of the terms is $2$.
Hence $$a_i = (i^2 - i + 1) + (i^2 - i + 3) + \ldots + (i^2 - i + 1 + 2(i-1))$$
Or, $$a_i = \sum_{j=1}^i \left(i^2 - i + (2j-1) \right)$$
Or, $$a_i = i^3 - i^2 + i^2 = i^3$$
Finally $$\sum_{i=1}^n a_i = \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$
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