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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

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  • $\begingroup$ you proved what was asked.then why do you sub in $n^2$, ? $\endgroup$ – sirous Feb 13 at 6:22
  • $\begingroup$ apologies I missed putting in the step I was working on. I'm now on the step "write the summation that gives the sum of the odd numbers found in the first n rows." of faulhabers formula which is given to the a_5 row. I have edited the post. $\endgroup$ – user886969 Feb 13 at 6:30
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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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  • $\begingroup$ This makes sense, thank you! I was spinning my wheels on how to make this logical jump. $\endgroup$ – user886969 Feb 13 at 7:14
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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 $$ \sum_{k=1}^{\text {final}}(2k-1) = ({\text {final}})^2. $$

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