How to calculate a combinatorial sum by hand My question is if there is an "easy" way to calculate
$$
\displaystyle \sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k}
$$
I know that if it were just $\displaystyle \sum_{k=0}^{6}\binom{6}{k}(2)^k(3)^{6-k}$, this is the same as $(2+3)^6 = 5^6 = 15625$. But with that $2n$ term in there, I'm not sure if there is some clever trick to still do something like $(2+3)^6$. Thanks for the help!

You can write this as
\begin{align}
\sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k} &= 2\cdot3^6 \sum_{k=0}^{6}k\binom{6}{k}(2)^k(3)^{-k} \\
&= 2\cdot3^6 \sum_{k=0}^{6}k\binom{6}{k} \left(\frac{2}{3}\right)^k, \\
\end{align}
and use the identity
$$\sum_{k = 0}^n k \binom{n}{k} x^k = x \frac{d}{dx} (1 + x)^n = nx(1+x)^{n-1}. $$
This gives you the answer
$$ \sum_{k=0}^{6}(2k)\binom{6}{k}(2)^k(3)^{6-k} = 2 \cdot 3^6 \cdot 6 \cdot \frac{2}{3} \left( 1 + \frac23 \right)^5 = 75000.$$
The general formula, which you can prove in the same way, is
$$ \sum_{k = 0}^n k \binom{n}{k}x^ky^{n - k} = nx(x + y)^{n - 1}. $$
