Consecutive product sum [Solution]

The task was to find (without pen and paper)

$$\sum_{k=r}^{n} k(k-1)\dots(k-r)$$

Solution

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Let $S_k = k(k-1) \dots (k-r)$.

One way to solve this is to notice that

$$(k+1)k(k-1)\dots(k-r) - k(k-1)(k-2)\dots(k-r)(k-r+1)$$

$$= (k+1-(k-r+1))k(k-1)\dots(k-r) = r k(k-1)\dots(k-r)$$

i.e

$$P_k - P_{k-1} = rS_k$$

where $P_k = (k+1)k(k-1) \dots (k-r)$

Thus we get a sum where most of the terms cancel each other out (called a telescopic sum) and the answer is simply $P_n - P_{r-1}$ which is

$$\frac{(n+1)n(n-1)\dots(n-r)}{r}$$

There are other solutions, like writing in terms of binomial coefficients etc, not sure we can do those without pen and paper though.