1
$\begingroup$

Find a formula for $\sum_{k=-3}^n$ $k^3 (k + 1) (k + 2) (k + 3)$ and prove it.

I'm pretty sure that once I find a formula I can prove it by induction. I'm just struggling to find the formula. My professor hasn't really taught us how to do that part. I can plug numbers in to get an idea, but I can't seem to formulate a formula. Suggestions?

For $n=1$, the result is 24. For $n = 2$, the result is 504. I'm guessing the formula will involve factorials.

$\endgroup$
2
  • 1
    $\begingroup$ See Faulhaber's formula. $\endgroup$
    – John Omielan
    Oct 9, 2020 at 1:08
  • 1
    $\begingroup$ Re John Omielan's comment, the problem can not be attacked without attacking Bernoulli numbers. If you are able to grasp the ideas, and highjack the first few Bernoulli numbers and apply them, then you are (in effect) done. If not, one idea is to acquaint yourself with the Calculus oriented definition of Bernoulli numbers, and generate the list yourself. If you wish to avoid all Calculus, and derive the necessary formulas yourself, purely through algebra, see math.stackexchange.com/questions/2697950/…. $\endgroup$
    – user2661923
    Oct 9, 2020 at 1:22

2 Answers 2

Reset to default
2
$\begingroup$

Here’s a method of solving it. It will take a little bit of work (algebra), but you should be able to do it.

Let $S(n)$ be the value of this sum, in terms of the upper limit $n$. Then you know that

$$\begin{align}S(n) - S(n-1) &= n^3(n+1)(n+2)(n+3) \\&= n^3(n^3 + 6n^2 + 11n + 6) \\&= n^6 + 6n^5 + 11n^4 + 6n^3\end{align}$$

There are infinitely many polynomials in $n$ satisfying this difference equation, and they differ by a constant. It turns out that $S(n)$ can be uniquely expressed as one of these seventh-degree polynomials:

$$S(n)=x_7n^7 + x_6 n^6 + ...+x_1 n + x_0$$

You just need to figure out the values of these eight coefficients. To do this, you can treat the following sequence of identities as a system of eight equations with eight unknowns $x_i$:

$$0x_7 + 0x_6 + ... + 0x_1 + 1x_0 = S(0) \\ 1x_7 + 1x_6 + ... + 1x_1 + 1x_0 = S(1) \\ 2^7 x_7 + 2^6 x_6 + ... + 2x_1 + 1x_0 = S(2) \\ ... \\ 7^7 x_7 + 7^6 x_6 + ... + 7x_1 + 1x_0 = S(7)$$

You can solve this system easily with a bit of linear algebra, but you probably don’t want to do it by hand. And once you find out the values of the $x_i$, you will know the coefficients of the polynomial $S(n)$.

$\endgroup$
2
$\begingroup$

Note $$ \begin{align*} k^3 &= k^3-1+1 \\ &= k(k-1)(k+1) + 1 \\ &= k(k-1)(k-2) + 3k(k-1) + 1 \end{align*} $$

So given is $$ \sum (k-2)(k-1)k(k+1)(k+2)(k+3) + 3{\sum(k-1)k(k+1)(k+2)(k+3)} + \sum (k+1)(k+2)(k+3) $$

$$ = (1\cdot2\cdot3\cdot4\cdot5\cdot6 + 2\cdot3\cdot4\cdot5\cdot6\cdot7 + \ldots) + 3(1\cdot2\cdot3\cdot4\cdot5 + 2\cdot3\cdot4\cdot5\cdot6 + \ldots) + (1\cdot2\cdot3 + 2\cdot3\cdot4 + \ldots) $$

Can you end it now?

(Careful about the new bounds)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.