Is there an equation for the sum of alternating cubes?

Question

The following sequences for sum of alternating cubes:

Odd cubes: [1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600]

Even cubes: [8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800]

Nothing on OEIS. Trying to find equations for generating such values.

Answer 1

There is a very general way for producing formulae for functions $F:\mathbb{N} \rightarrow \mathbb{R}.$ Let me explain.

Consider the set of functions $F:\mathbb{N} \rightarrow \mathbb{R}.$ On such functions there is a discrete analog of differentiation called the shift operator, denoted $\Delta$, which given a function $F$ produces a function $\Delta F$ via the formula $\Delta F(x) := F(x+1) - F(x).$ Now classically, if one is given an infinitely differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ one can consider the Taylor series $\displaystyle\sum_{i=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k$ and ask if this series converges and is equal to $f$ near $0.$ In the discrete case, note that iterated powers of the shift operator are always well defined. So it is natural to ask if there is a discrete analog of Taylor series. These analogs are called Newton series. Given a function $F:\mathbb{N} \rightarrow \mathbb{R},$ we define its Newton series via the formula

$$\sum_{k=0}^{\infty} \Delta^k F(0){{x}\choose{k}}.$$

The next question to ask is when is $F$ equal to its Mahler expansion. And this is where the discrete case is nicer than the classical, for the answer is always:

For all $x \in \mathbb{N}$, we have $F(x) = \sum_{k=0}^{\infty} \Delta F^k(0){{x}\choose{k}}$.

(Note if $k>x,$ we have ${x\choose k} = 0$ so there is no issue of convergence in the above series).

So if I have such a function $F:\mathbb{N} \rightarrow \mathbb{R}$ which I want to express via a nice formula, a good place to begin is to try to write down its Newton expansion.

Let's consider your case where

$$F(x) = \sum_{k=0}^{x} (2i)^3.$$

Then

$$\Delta F(x) = (2(x + 1))^3$$

$$\Delta^2 F(x) = (2(x + 2))^3 - (2(x + 1))^3 = 24x^2 +72x + 56$$

$$\Delta^3 F(x) = \Delta^2 F(x+1) - \Delta^2 F(x) = 192x + 208$$

$$\Delta^4 F(x) = \Delta^3 F(x+1) - \Delta^3 F(x) = 192$$

and

$$\Delta^k F(x) = 0 \text{ if } k\ge5.$$

It follows

$$F(x) = {{x}\choose{1}} + 56{{x}\choose{2}} + 208{{x}\choose{3}} + 192{{x}\choose{4}}.$$

Following the same mechanical procedure we can also produce a formula for the some of the cubes of the odd natural numbers and many other functions of this sort.

Answer 2

$$\sum\limits_{i=1}^n(2i)^3=8\sum\limits_{i=1}^ni^3$$ $$\sum\limits_{i=1}^n(2i-1)^3=\sum\limits_{i=1}^{2n}i^3-\sum\limits_{i=1}^n(2i)^3$$

Of course, the sum of cubes is (for some strange reason...) the square of the sum of consecutive integers.

$$\sum\limits_{i=1}^ni^3=\left(\sum\limits_{i=1}^ni\right)^2=\frac{n^2(n+1)^2}4$$

Answer 3

For the odd cubes, you are asking about $\sum_{i=0}^n (2i+1)^3=\sum_{i=0}^n 8i^3+12i^2+6i+1$. Now feed each term to Faulhaber's formula

For the even cubes, you want $\sum_{i=1}^n (2i)^3=8\sum_{i=1}^n i^3$