Changing limits of a summation?

I've been doing a question which asks you to prove the following is true: $$\sum_{r=1}^{n}r(r+1)(r+2)(r+3)(r+4)$$ $$=\frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)$$ which is realtively straightforward using induction. I'm wondering how you can use the above formula to conclude that: $$\sum_{r=0}^{n-1}r(r-1)(r-2)(r-3)(r-4)$$ $$=\frac{1}{6}n(n-1)(n-2)(n-3)(n-4)(n-5)$$

It seems intuitive that it should be true since we've just shifted things around a bit. But I can't seem to be able to show it directly, I've tried index shifts but I can't seem to get it; though it's definitely correct.

• Not sure if this will work, but try substituting k=r-4. Thus the terms in the bottom sum are identical to the ones in the top sum. Make sure to change the bounds of the sum, i.e. k=-4, k=n-5. Then you can find the value of the sum from k=-4 to k=-1, and add that value on. Can you figure out the sum from there? – Cataline Jun 18 '15 at 16:39

4 Answers

How about this? \begin{align}\frac{1}{6}(n-5)(n-4)(n-3)(n-2)(n-1)n&=\sum_{r=1}^{n-5}r(r+1)(r+2)(r+3)(r+4)\\&=\sum_{r=5}^{n-1}(r-4)(r-3)(r-2)(r-1)r\\&=\sum_{r=0}^{n-1}r(r-1)(r-2)(r-3)(r-4)\end{align}

First note that by eliminating the zero terms, we have $$\sum_{r=0}^{n-1}r(r−1)(r−2)(r−3)(r−4)=\sum_{r=5}^{n-1}r(r−1)(r−2)(r−3)(r−4).$$ Now let $k=r-4$. We have $$\sum_{r=5}^{n-1}r(r−1)(r−2)(r−3)(r−4)=\sum_{k=1}^{n-5}k(k+1)(k+2)(k+3)(k+4).$$ Apply the original identity and plug $r$ back in for the answer :)

You have $$\sum_{r=1}^{r=n}r(r+1)(r+2)(r+3)(r+4)=\frac{1}{6}n(n+1)(n+2)(n+3)(n+4)(n+5)$$ and thus $$\sum_{r=1}^{r=k-5}r(r+1)(r+2)(r+3)(r+4)=\frac{1}{6}(k-5)(k-4)(k-3)(k-2)(k-1)k$$ Writing $r=s-4$ (that implies $s=r+4$), we get

$$\sum_{s=5}^{s=k-1}(s-4)(s-3)(s-2)(s-1)s=\frac{1}{6}(k-5)(k-4)(k-3)(k-2)(k-1)k$$

But for $s=0,1,2,3,4$ we have $(s-4)(s-3)(s-2)(s-1)s=0$ so that

$$\sum_{s=0}^{s=k-1}(s-4)(s-3)(s-2)(s-1)s=\frac{1}{6}(k-5)(k-4)(k-3)(k-2)(k-1)k$$ Just changing the notation ($s$ by $r$, $k$ by $n$) and the order of the factors,

$$\sum_{r=0}^{r=n-1}r(r-1)(r-2)(r-3)(r-4)=\frac{1}{6}n(n-1)(n-2)(n-3)(n-4)(n-5)$$ as desired.

Original identity: $$\scriptsize\color{lightgrey}{\sum_{r=1}^nr(r+1)(r+2)(r+3)(r+4)=} \small5!\sum_{r=1}^n \binom {r+4}5=5!\binom {n+5}6\color{lightgrey}{=\scriptsize\frac 16n(n+1)(n+2)(n+3)(n+4)(n+5)}$$ Put $m-1=n$ and $j=r+4$: $$5!\sum_{r=1}^{m-1} \binom {j}5=5!\binom {m}6\qquad\quad$$ Replacing $m$ with $n$, and $j$ with $r$: $$\scriptsize\color{lightgrey}{\sum_{r=1}^nr(r-1)(r-2)(r-3)(r-4)=}\small5!\sum_{r=1}^{n-1} \binom {r}5=5!\binom {n}6\small\color{lightgrey}{=\scriptsize\frac 16n(n-1)(n-2)(n-3)(n-4)(n-5)}$$ as required.