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!
 A: This approach avoids differentation. Use the identitiy
$$
\binom{n}{k}\binom{k}{m}=\binom{n}{m}\binom{n-m}{k-m};\quad n\geq k\geq m.\tag{1}
$$
Note we may start the summation at $k=1$ (first term in sum is zer0). Then
$$
\sum_{k=1}^{6}(2k)\binom{6}{k}2^k(3)^{6-k}
=2\sum_{k=1}^{6}6\binom{5}{k-1}2^k(3)^{6-k}
=2(6)(2)\sum_{u=0}^{5}\binom{5}{u}2^u(3)^{5-u}\tag{2}
$$
where we used (1) in the first equality and made a change of variables ($u=k-1$) in the second equality. We can compute the last sum by the binomial theorem. Indeed,
$$
\sum_{k=0}^{6}(2k)\binom{6}{k}2^k(3)^{6-k}=24(2+3)^5=75000.\tag{3}
$$
A: 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
$$ \sum_{k = 0}^n k \binom{n}{k}x^ky^{n - k} = nx(x + y)^{n - 1}. $$
A: Another approach is to try to interprate the sum as a way of counting the number of ways that something can happen. In this case, consider the following scenario:
We wish to start with a group of six people, give each person one of five different types of fruit in such a way that at least one person gets either an apple or a banana, and pick one person who got either an apple or a banana as the leader of the group.
One way that we can count the number of ways that this can be done is to count it based on the number of people who got either an apple or a banana. If $k$ people got either an apple or a banana, then there are $\binom{6}{k}$ ways to choose which $k$ people these are. There are then $k$ ways to choose who is the leader of the group. For each choice of leader, there are $2^k$ ways to pick the specific fruit that each of the $k$ people got (each got either an apple or a banana), and there are $3^{6-k}$ ways to choose what fruit each of the remaining people got. (There are $3$ options for each of the $6 - k$ remaining people.)
Thus there are
$$
  k \cdot \binom{6}{k} \cdot 2^{k} \cdot 3^{6 - k}
$$
possibilities if there are $k$ people who got either an apple or a banana. In total, this gives us
$$
  \sum_{k=1}^{6} k \cdot \binom{6}{k} \cdot 2^{k} \cdot 3^{6 - k}
$$
possibilities. This is precisely half of the sum which you wished to evaluate.
The other way that we can count the number of possibilities is to first pick the leader. There are $6$ ways in which this can be done. There are then two options for which fruit the leader got. (Either an apple or a banana.) We can then give all of the remaining people one of five different fruit in $5^5$ ways. This gives us a total of $6 \cdot 2 \cdot 5^5 = 37500$ ways that we can carry out the procedure, and so we have that
$$
  \sum_{k=1}^{6} k \cdot \binom{6}{k} \cdot 2^{k} \cdot 3^{6 - k} = 37500.
$$
The sum that you wanted is double this, and hence is equal to $75000$.
