Problem source

Write down the binomial expansion of $(1+x)^{n}$, where $n$ is a
positive integer. 
\begin{questionparts}
\item By substituting particular values of $x$ in the above expression, or otherwise, show that, if $n$ is an even positive integer, 
\[
\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{n}=\binom{n}{1}+\binom{n}{3}+\binom{n}{5}+\cdots+\binom{n}{n-1}=2^{n-1}.
\]
\item Show that, if $n$ is any positive integer, then 
\[
\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n2^{n-1}.
\]
\end{questionparts}
Hence evaluate 
\[
\sum_{r=0}^{n}\left(r+(-1)^{r}\right)\binom{n}{r}\,.
\]

Solution source

\[ (1+x)^n = \sum_{k=0}^n \binom{n}{k} x^k \]

\begin{questionparts}
\item \begin{align*}
(1+1)^n &= \sum_{k=0}^n \binom{n}{k} \\
(1-1)^n &= \sum_{k=0}^n (-1)^n\binom{n}{k} \\
&= \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} -\sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k} 
\end{align*}

Therefore $\displaystyle \sum_{\text{even }k, 0 \leq k \leq n} \binom{n}{k} = \sum_{\text{odd }k, 0 \leq k \leq n} \binom{n}{k}  = \frac{2^n}{2} = 2^{n-1}$

\item \begin{align*}
&& (1+x)^n &= \sum_{k=0}^n \binom{n}{k}x^k \\
\frac{\d}{\d x}: && n(1+x)^{n-1} &= \sum_{k=0}^n k\binom{n}{k} x^{k-1} \\
x = 1: && n2^{n-1} &= \sum_{k=1}^n k\binom{n}{k}
\end{align*} as required
\end{questionparts}

\begin{align*}
\sum_{r=0}^n (r + (-1)^r) \binom{n}{r} &= n2^{n-1}+0 = n2^{n-1}
\end{align*}