Summation using binomial coefficient I am trying to calculate the following using binomial coefficients and summation, but my memory is standing in the way:
$$
   \sum_{k=1}^n {k} * 2 ^ {k - 1}
$$
Thanks!
With great help I got to:
$$ (n - 1)* 2 ^{n} - 1 $$
Can you please confirm this?
 A: Hint: switch to polynomial functions.
$$
 P(x)=\sum_{k=1}^n kx^{k-1}
$$
A: Clement C's solution is the best approach. There is another approach, however.
Notice that
$$
\sum_{k=1}^n k2^{k-1} = \sum_{k=1}^n \sum_{i=1}^k 2^{k-1}
$$
because $k=\sum_{i=1}^k 1$. By reversing the order of summation, you get
$$
\sum_{k=1}^n k2^{k-1} = \sum_{i=1}^n \sum_{k=i}^n 2^{k-1}
$$
A closed form solution can now be found easily for the inner sum, which will produce a form that can be used to find a closed form solution for the outer sum.
A: Here is another approach, using
$$
\sum_{k=0}^n\binom{n}{k}=2^n
$$
and
$$
\sum_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}
$$
to get
$$
\begin{align}
\sum_{k=1}^nk\,2^{k-1}
&=\sum_{k=1}^nk\,\sum_{j=0}^{k-1}\binom{k-1}{j}\\
&=\sum_{j=0}^{n-1}\sum_{k=j+1}^n(j+1)\binom{k}{j+1}\\
&=\sum_{j=0}^{n-1}(j+1)\binom{n+1}{j+2}\\
&=\sum_{j=1}^nj\binom{n+1}{j+1}\\
&=\sum_{j=1}^n(j+1)\binom{n+1}{j+1}-\binom{n+1}{j+1}\\
&=\sum_{j=1}^n(n+1)\binom{n}{j}-\binom{n+1}{j+1}\\
&=(n+1)(2^n-1)-(2^{n+1}-n-2)\\[12pt]
&=(n-1)2^n+1
\end{align}
$$

Verification
Of course, to verify, we can try induction.
The formula works for $n=1$: $1=1$.
Assume the formula works for $n-1$, then
$$
\begin{align}
\sum_{k=1}^nk\,2^{k-1}
&=n\,2^{n-1}+\sum_{k=1}^{n-1}k\,2^{k-1}\\
&=n\,2^{n-1}+(n-2)2^{n-1}+1\\[9pt]
&=(n-1)2^n+1
\end{align}
$$
Thus, it is true for all $n\ge1$.
A: Here is another approch
∑k.2^(k-1)
Let f(x)=∑k.x^(k-1)
∫f(x) dx=∑∫k.x^(k-1)dx
∫f(x) dx=∑x^k
∫f(x)dx= x +x^2 + x^3 + …………………. X^n
∫f(x) dx= x(x^n-1)/(x-1)
Differentiating both sides 
f(x)= ( nx^(n+1) –(n+1)x^n + 1)/(x-1)^2
put x=2  f(2)=(n-1)2^n +1
