I am unsure of how Wolfram Alpha computes the following partial sum:
sum of (n-k)!/(n-r-k)! from 1 to n-r - Wolfram|Alpha
If someone could explain to me how the equation for the partial sum is derived I'd greatly appreciate it. Thanks!
I am unsure of how Wolfram Alpha computes the following partial sum:
sum of (n-k)!/(n-r-k)! from 1 to n-r - Wolfram|Alpha
If someone could explain to me how the equation for the partial sum is derived I'd greatly appreciate it. Thanks!
First, simplify...
$\displaystyle n - k - r < n - k$, so
$\displaystyle (n - k)! = (n - k)(n - k - 1)(n - k - 2)\dots (n - k - r) \dots 3 \cdot 2 \cdot 1$
$\displaystyle = (n - k)(n - k - 1)(n - k - 2)\dots (n - k - r)!$.
Therefore
$\displaystyle \frac{(n - k)!}{(n - k - r)!} = \frac{(n - k)(n - k - 1)(n - k - 2)\dots (n - k - r)!}{(n - k - r)!}$
$\displaystyle = (n - k)(n - k - 1)(n - k - 2) \dots (n - k - r + 1)$.
Can you now try to work out
$\displaystyle \sum_{k = 1}^{n - r} (n - k)(n - k - 1)(n - k - 2) \dots (n - k - r + 1)$?