Question

Evaluate the following sum:

Answer 1

\[\sum_{k=0}^{n} k(n-k)\left(\begin{matrix}n \\ k\end{matrix}\right)\]

Answer 2

use the definition of \({n\choose k}\) to simplify

Answer 3

and the fact that

\[\sum_{k=0}^{m}{m\choose k}=2^m\]

Answer 4

@zarkon, does it simplify to
\[\sum \frac{n!}{(k-1)!(n-k-1)!}\] or am i way off?

Answer 5

yes

Answer 6

also notice that

\[\sum_{k=0}^{n} k(n-k)\left(\begin{matrix}n \\ k\end{matrix}\right)\]
\[=\sum_{k=1}^{n-1} k(n-k)\left(\begin{matrix}n \\ k\end{matrix}\right)\]

Answer 7

\(n\) is fixed right? so is this the same as
\[n\sum\dbinom{n-1}{k-1}\]?

Answer 8

n is fixed...but you are a little off

Answer 9

\[n-1-(k-1)=n-k\ne n-k-1\]

Answer 10

oh yes a am off

Answer 11

just factor out n-1

Answer 12

in addition to the n you already factored out

Answer 13

\[n(n-1)\sum_{}^{} \left(\begin{matrix}n-2 \\ k-1\end{matrix}\right) \ \ \ \ ?\]

Answer 14

yes

Answer 15

let \(w=k-1\)

then you get

\[n(n-1)\sum_{w=0}^{n-2}{n-2\choose w}\]

Answer 16

the ans is \[n(n-1)2^{n-2} \]:)
Thank you!

Answer 17

yep