Find a closed expression for a formula including summation Let:
$$\sum\limits_{k = 0}^n {k\left( {\matrix{
   n  \cr 
   k  \cr 
 } } \right)}  \cdot {4^{k - 1}} \cdot {3^{n - k}}$$
Find a closed formula (without summation). I think I should define this as a "series" which generated by $F(x)$. I don't really have a lead here.
Any ideas? Thanks. 
 A: Hint: Use the fact that for $k\ge 1$ we have $k\binom{n}{k}=n\binom{n-1}{k-1}$ and think Binomial Theorem. 
A: As a Hint :
\begin{align*}
\sum_{k = 0}^n k\binom{n}{k}4^{k-1}\cdot3^{n-k}=0\cdot\binom{n}{0}\cdot4^{-1}\cdot3^n+\sum_{k = 1}^n n\binom{n-1}{k-1}4^{k-1}\cdot 3^{n-k}
\end{align*}
A: Hint:
$$\sum\limits_{k = 0}^n {\left( {\matrix{
   n  \cr 
   k  \cr 
 } } \right)}  \cdot {x^{k}} \cdot {3^{n - k}}=\left(x+3\right)^n$$
A: The binomial expansion is given by
\begin{align}
(1+x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}.
\end{align}
Differentiating both sides with respect to $x$ yields
\begin{align}
n (1+x)^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} x^{k-1}.
\end{align}
Multiplying this last expression by $3^{n} 4^{-1}$ leads to
\begin{align}
\frac{3^{n} n}{4} \ x(1+x)^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} 4^{-1} 3^{n} x^{k}.
\end{align}
Now let $x = 4/3$ to obtain the desired expression
\begin{align}
 n \cdot 7^{n-1} = \sum_{k=0}^{n} k \ \binom{n}{k} 4^{k-1} 3^{n-k}.
\end{align}
A: According to the binomial theorem, we have
$$
(x+y)^n=\sum_{k=0}^n\binom{n}{k} x^k y^{n-k}.\tag1
$$
Differentiating $(1)$ with respect to $x$ yields
$$
n(x+y)^{n-1}=\sum_{k=0}^n \binom{n}{k} k\ x^{k-1} y^{n-k},\tag2
$$
then pluging in $x=4$ and $y=3$ to $(2)$.
