# 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

• you proved what was asked.then why do you sub in $n^2$, ? Commented Feb 13, 2021 at 6:22
• 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.
– user886969
Commented Feb 13, 2021 at 6:30

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 $$\sum_{i=1}^n a_i = \sum_{i=1}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2$$

• This makes sense, thank you! I was spinning my wheels on how to make this logical jump.
– user886969
Commented Feb 13, 2021 at 7:14

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 $$\sum_{k=1}^{\text {final}}(2k-1) = ({\text {final}})^2.$$