Proof of equation with binomial coefficients: $\sum\limits_{k=1}^{n} (k+1) \binom{n}{k} = 2^{n-1} \cdot (n+2)-1$ $$\sum\limits_{k=1}^{n} (k+1) \binom{n}{k} = 2^{n-1} \cdot (n+2)-1$$
Maybe it's simple to prove this equation but I'm not sure how to get along with the induction. Any hints for this? Or may I use another method?
Thanks a lot!
 A: It isn’t quite right as stated. Try $n=1$, for instance: the lefthand side is
$$1\binom10+2\binom11=3=2^0\cdot3\;,$$
not $2^0\cdot3-1$. You can get the correct identity easily from the identity $k\binom{n}k=n\binom{n-1}{k-1}$:
$$\begin{align*}
\sum_{k=0}^n(k+1)\binom{n}k&=\sum_{k=0}^nk\binom{n}k+\sum_{k=0}^n\binom{n}k\\
&=n\sum_{k=0}^n\binom{n-1}{k-1}+2^n\\
&=n\sum_{k=0}^{n-1}\binom{n-1}k+2^n\\
&=2^{n-1}n+2^n\\
&=2^{n-1}(n+2)\;.
\end{align*}$$
Added: A combinatorial proof is also possible. You have a pool of $n$ men and a woman. From this pool you are to form a committee of any size, except that it must include the woman, and to appoint one member of the committee to be chair. If the committee has $k$ men, where $0\le k\le n$, there are $\binom{n}k$ ways to choose them, and there are then $k+1$ ways to choose the chair of the committee. Summing over the possible values of $k$, we see that there are
$$\sum_{k=0}^n(k+1)\binom{n}k$$
ways to form the committee and choose its chair.
Alternatively, we can pick the chair first. If we pick one of the $n$ men to be chair, we can then select any subset of the remaining $n-1$ men and form the committee from these men and the woman; this can be done in $n2^{n-1}$ ways. If we pick the woman to be chair, we can fill out the committee with any of the $2^n$ subsets of the pool of men. Thus, there are $n2^{n-1}+2^n$ ways to form the committee and choose its chair, and the result follows.
Added2: Since the $k=0$ term on the lefthand side is $1$, the identity could also be corrected to
$$\sum_{k=1}^n(k+1)\binom{n}k=2^{n-1}(n+2)-1\;.$$
A: Another proof (without calculus).
$$\sum\limits_{k=0}^{n} (k+1) \binom{n}{k} = \sum\limits_{k=0}^{n} k\binom{n}{k}+\sum\limits_{k=0}^{n} \binom{n}{k}\\
=\sum\limits_{k=1}^{n} n\binom{n-1}{k-1}+2^n
=n2^{n-1}+2^n=2^{n-1} \cdot (n+2)$$
which is a bit different from your formula.
A: $$\sum_{k=0}^{n} (k+1) \binom{n}{k} =\sum_{k=0}^{n}k\binom{n}{k}+\sum_{k=0}^{n} \binom{n}{k}=\sum_{k=0}^{n}k\cdot\frac{n}{k}\binom{n-1}{k-1}+\sum_{k=0}^{n} \binom{n}{k}=$$
$$=n\sum_{k=0}^{n}\binom{n-1}{k-1}+\sum_{k=0}^{n}\binom{n}{k}=$$
$$=n\sum_{k=1}^{n}\binom{n-1}{k-1}+\sum_{k=0}^{n}\binom{n}{k}=n2^{n-1}+2^n $$
A: Start with
$$
(1 + x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k}.
$$
Multiply by $x$ to get
$$
x (1 + x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{k+1}.
$$
Differentiate to get
$$
(1 + x)^{n} + n x (1 + x)^{n-1} = \sum_{k=0}^{n} (k + 1) \binom{n}{k} x^{k}.
$$
Set $x = 1$ to get
$$
2^{n} + n 2^{n-1} = \sum_{k=0}^{n} (k + 1) \binom{n}{k},
$$
which is precisely your identity, except for the final $-1$ which I believe to be incorrect.
A: $$k\binom nk=k\frac{n!}{k!(n-k)!}=\frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}=n\binom{n-1}{k-1}.$$
Then the original summation will yield terms $n2^{n-1}$ and $2^n$, or $2^{n-1}(n+2)$.
Note that the original sum has $n+1$ terms, while the transformed one has only $n$. But this makes no difference as the first term is with $k=0$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 1}^{n}\pars{k + 1}{n \choose k} & =
-1 + \sum_{k = 0}^{n}\pars{k + 1}{n \choose k}
\\[5mm] & =
-1 + {1 \over 2}\sum_{k = 0}^{n}\bracks{%
\pars{k + 1}{n \choose k} + \pars{n - k + 1}{n \choose n - k}}
\\[5mm] & =
-1 + {1 \over 2}\pars{n + 2}\sum_{k = 0}^{n}{n \choose k} =
-1 + {1 \over 2}\pars{n + 2}\,2^{n} =
\bbx{\ds{2^{n - 1}\pars{n + 2} - 1}}
\end{align}
