# Finding closed form for $1^3+3^3+5^3+…+n^3$

I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.

How would I go about doing this?

I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.

My gut feeling is telling me to multiply the above series by 8, then subtract it from the original, but it doesn't quite get me there because I'm going to have a whole lot of extra terms that I do not want.

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By factorizing $2^3 = 8$ from the even terms, we can re-express the sum of the even terms:

$$2^3 + 4^3 + \dots + (n-1)^3 = 8\left(1^3 + 2^3 + \dots + \left(\frac{n-1}{2}\right)^3\right)$$

Simply subtract this from $1^3 + 2^3 + \dots + n^3$ to give the sum of the remaining odd terms. You can compute the formula for the RHS of the above equation using the equation you mentioned.

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It may be simpler to use a method that does not require any creativity:

Let n = (2k-1) here.

$\sum_{i=1}^k(2i-1)^3 = \sum_{i=1}^k(8i^3 - 12i^2 + 6i - 1) = 8\sum_{i=1}^ki^3 - 12\sum_{i=1}^ki^2 + 6\sum_{i=1}^ki - \sum_{i=1}^k1 = 8(\frac{k(k+1)}{2})^2 - 12\frac{k(k+1)(2k+1)}{6} + 6\frac{k(k+1)}{2} - k$

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Be aware, though, that $\sum_{i=0}^k (2i+1)^3$ is the expression OP wants to find, not $\sum_{i=1}^k (2i+1)^3$. – user26486 Jun 26 '14 at 12:08
Right, edited it. – Wonder Jun 26 '14 at 12:34

Using repeated differences we get $$\begin{array}{llll} 1 & 28 & 153 & 496 & 1225 & 2556 & \\ 27 & 125 & 343 & 729 & 1331 & \\ 98 & 218 & 386 & 602 & \\ 120 & 168 & 216 & \\ 48 & 48 & \\ 0 & \\ \end{array}$$ Newton's interpolation formula then gives us $$1 \binom{n-1}{0} + 27 \binom{n-1}{1} + 98 \binom{n-1}{2} + 120 \binom{n-1}{3} + 48\binom{n-1}{4} = 2 n^4-n^2$$

This is a technique well worth knowing. (The simplification was done with WA.)

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Note that $$\sum 1=n=C_n^1$$ $$\sum\sum1=\sum i=\frac{n(n+1)}{2}=C_{n+1}^2$$ $$\sum\sum\sum1=\sum\frac{i(i+1)}2=\frac{n(n+1)(n+2)}{2.3}=C_{n+2}^3$$$$\sum\sum\sum\sum1=\sum\frac{i(i+1)(i+2)}{2.3}=\frac{n(n+1)(n+2)(n+3)}{2.3.4}=C_{n+3}^4.$$

These are diagonals in Pascal's triangle.

From there, you derive $$\sum i^2=2\sum\frac{i(i+1)}2-\sum i=2C_{n+2}^3-C_{n+1}^2,$$ and $$\sum i^3=6\sum\frac{i(i+1)(i+2)}{6}-3\sum i^2-2\sum i=6C_{n+3}^4-6C_{n+2}^3+C_{n+1}^2.$$

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If

$\color{blue}{n = 1}$, $\sum$ = 1

$\color{blue}{n=2}$, $\sum = 1^3 + 3^3 = 28 = 2^2 \cdot 7 = \color{blue}{2}^2(2\cdot \color{blue}{2}^2-1)$

$\color{blue}{n = 3}, \sum = 1^3 + 3^3 + 5^3 = 153 = 3^2 \cdot 17 = \color{blue}{3}^2(2\cdot \color{blue}{3}^2 - 1)$

$\color{blue}{n = 4}, \sum = 1^3 + 3^3 + 5^3 + 7^3 = 496 = 2^4\cdot 31 = \color{blue}{4}^2(2\cdot \color{blue}{4}^2 - 1)$

So, we can suppose that $$\sum_{k=1}^{n} (2k-1)^3 = n^2(2n^2 - 1)$$ This formula can be proved by method of mathematical induction.

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Just for the collection:

From the identity $(2k-1)^3 = 2k^4-k^2 - 2(k-1)^4+(k-1)^2$, we see that if we denote $u_k = 2k^4-k^2$ then we have $(2k-1)^3=u_k - u_{k-1}$.

Thus, $$1^3+3^3+\cdots+(2n-1)^3 = \sum_{k=1}^n (2k-1)^3 = \sum_{k=1}^n (u_k - u_{k-1}) = u_n-u_0 = 2n^4-n^2.$$

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