Sum of a series problem involving cubes

Question

For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$

How would you go about solving such a problem? Any help would be appreciated.

Answer 1

I would use the following ideas:

(1): Standard formula: $1^3+2^3+\cdots+a^3=\left(\frac{a(a+1)}{2}\right)^2$

(2): Use (1) to get an expression for $2^3+4^3+6^3+\cdots+(2a)^3=2^3\cdot(1^3+2^3+\cdots+a^3)$

(3): Use (1) and (2) to get an expression for $1^3+3^3+5^3+\cdots+(2a-1)^3$ by noting that $$1^3+3^3+5^3+\cdots+(2a-1)^3=\left(1^3+2^3+\cdots+(2a)^3\right)-\left(2^3+4^3+\cdots+(2a)^3 \right)$$

(4) Your expression breaks down to an expression from (3) minus an expression from (2) (since your $n$ is supposed to be odd--if your $n$ were even you could still use these ideas to get your expression for the alternating sum.)

Answer 2

Just split the problem up into "odd" and "even" cases and use Faulhaber's formula (this is perhaps the most straightforward way); e.g.:

$$\begin{align} 7^3-6^3+5^3-4^3+3^3-2^3+1^3 \\ &= (7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\ &=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2(6^3+4^3+2^3) \\ &=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-2((2^3)(3^3+2^3+1^3)) \\ &=(7^3+6^3+5^3+4^3+3^3+2^3+1^3)-16(3^3+2^3+1^3) \end{align}$$

Alternatively, notice that if we define:

$$A_n = \sum_{k=1}^n{((-1)^{n-k} \cdot k^3)}$$

for the terms in a sequence, then:

$$A_n + A_{n+1} = (n+1)^3$$

It goes kind of like this: Define $f(x)$ to be:

$$f(x) = \sum_{k=0}^\infty{(A_kx^k)}$$

Then, calculate $(x+1)f(x)=xf(x)-f(x)$ (as a rational function) by applying:

$$A_n + A_{n+1} = (n+1)^3$$

This gives the generating function of $f(x)$ (it is a rational function). Expand it using partial fractions; each partial fraction is a geometric series or a power (i.e., derivative) of a geometric series.

Express each partial fraction as a simple series, add them, and equate coefficients to $A_n$. And you get the answer!!!

See the sequence at OEIS, for example.

The answer is:

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

Answer 3

Considering the first four terms, we have

$$n^3-(n-1)^3+(n-2)^3-(n-3)^3+\dots$$

Finite difference methods suggest that $f(n)=n^3$ and here we have $f'(n-1)+f'(n-3)$ where $f'(n)=f(n+1)-f(n)$ is taken as the forward difference at $n$. The value of $f'(n)$ is easily determined as $f'(n)=3n^2+3n+1$. Looking closely, it is clear that there is an odd number of terms and so all pairs but the final $1$ in the sum reduce to this form. Since $n$ is odd we can reduce the question by $2k-1=n$ to

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

Reducing to use $n$ in place of $k$ should be a matter of relatively simple algebra.

Answer 4

The sums $S_n=\sum\limits_{k=1}^n(-1)^{n-k}k^3$ cannot coincide with the values of a polynomial $an^3+bn^2+cn+d$ for some given $(a,b,c,d)$, for every integer $n\geqslant1$. However, for every $n\geqslant0$,

$$S_n=\frac12n^3+\frac34n^2-\frac18+(-1)^n\frac18.$$

Thus, both sequences $(S_{2n})$ and $(S_{2n+1})$ coincide with the values of a polynomial of degree $3$, but not the same.

---

To prove this, one approach is to solve the identity $$n^3=P(n)+P(n-1),$$ for the polynomial $P(n)=an^3+bn^2+cn+d$. This yields $(a,b,c,d)=(\frac12,\frac34,0,-\frac18)$ hence $$P(n)=n^3-P(n-1)=n^3-(n-1)^3+P(n-2)=\cdots=S_n+(-1)^nP(0),$$ that is, $$S_n=P(n)-(-1)^nP(0).$$ Considering only odd integers $n$ yields $$S_n=\frac12n^3+\frac34n^2-\frac14.$$ that is, $$S_{2k+1}=4k^3+9k^2+6k+1.$$ Considering only even integers $n$ yields $$S_n=\frac12n^3+\frac34n^2,$$ that is, $$S_{2k}=4k^3+3k^2.$$