Compute $\sum_{k=1}^n \binom nk k 3^k$ I'm trying to compute
$$\sum_{k=1}^n \binom nk k 3^k$$
but don't know how. Would anyone be able to show me?
The only thing that I can possibly think of is that 
$$\sum_{k=1}^n \binom nk k 3^k = \frac{1}{\ln 3}\sum_{k=1}^n \binom nk \frac{d}{dk}\left[3^k\right]$$
Thanks
 A: $$\begin{array}{rcl}
\displaystyle \sum_{k=1}^n \binom nk k 3^k
&=& \displaystyle \sum_{k=1}^n \frac{n!}{(n-k)!k!} k 3^k \\
&=& \displaystyle \sum_{k=1}^n \frac{n!}{(n-k)!(k-1)!} 3^k \\
&=& \displaystyle \sum_{k=1}^n n \frac{(n-1)!}{(n-k)!(k-1)!} 3^k \\
&=& \displaystyle \sum_{k=1}^n n \binom{n-1}{k-1} 3^k \\
&=& \displaystyle n \sum_{k=1}^n \binom{n-1}{k-1} 3^k \\
&=& \displaystyle n \sum_{k=0}^{n-1} \binom{n-1}{k} 3^{k+1} \\
&=& \displaystyle 3n \sum_{k=0}^{n-1} \binom{n-1}{k} 3^k \\
&=& \displaystyle 3n (1+3)^{n-1} \\
&=& \displaystyle 3n \cdot 4^{n-1} \\
\end{array}$$
A: Hint: Differentiate $(1+x)^{n}$.
A: Hint: Your differentiation idea is a good one. Try writing
$$f(x) = \sum_{k=1}^n {n \choose k} k x^k$$
(so your aim is to calculate $f(3)$). Then notice that
$$f(x) = \sum_{k=1}^n {n \choose k} \left(x \cdot\frac{d}{dx} (x^k)\right).$$
A: $(1+x)^n = \sum_{k=0}^{n}\binom{n}{k}x^k;$
$x\frac{d}{dx} (1+x)^n= \sum_{k=0}^{n}\binom{n}{k}kx^k.$
$xn(1+x)^{n-1} = \sum_{k=1}^{n}\binom{n}{k}kx^k.$
A: You can write the binomial coefficient as:
$$\sum_{k=1}^n \binom nk k 3^k = 3n\sum_{k=1}^{n} \binom{n-1}{k-1} 3^{k-1}$$
Now note the binomial expansion of $(1+3)^{n-1}$
