TITLE: Solve summation expression
QUESTION [7 upvotes]: For a probability problem, I ended up with the following expression
$$\sum_{k=0}^nk\ \binom{n}{k}\left(\frac{2}{3}\right)^{n-k}\left(\frac{1}{3}\right)^k$$
Using Mathematica I've found that the result should be $\frac{n}{3}$. However, I have no idea how to get there. Any ideas?

REPLY [6 votes]: Just some binomial coefficient massage and the binomial formula:
$$
\sum_{k=0}^nk\binom nkx^{n-k}y^k
=\sum_{k=1}^nn\binom{n-1}{k-1}x^{n-k}y^k
=ny(x+y)^{n-1}
$$
Now plug in $x=\frac23$ and $y=\frac13$.