How to solve a summation problem involving binomial coefficient and powers? The problem is as follows:
$\textrm{Find the value of n in the summation:}$
$$\sum_{r=0}^{n}(2r+1)\binom{n}{r}=2^{n+4}$$
So far I only know the definition of the binomial coefficient as this:
$\binom{n}{k}=\frac{n!}{r!(n-k)!}$
But if I replace the above in the problem ends like this:
$$\sum_{r=0}^{n}(2r+1)\binom{n}{r}=\sum_{r=0}^{n}(2r+1)\frac{n!}{r!(n-r)!}$$
From then on I'm lost. How to solve that situation?.
 A: You can observe that, by Newton's formula
$$\sum_{r=0}^n \binom{n}{r} = (1+1)^n = 2^n$$
and so you can impose
$$2^{n+4} = 16 \sum_{r=0}^n \binom{n}{r} = \sum_{r=0}^n (2r+1) \binom{n}{r}.$$
Thus
$$\sum_{r=0}^n (15-2r)\binom{n}{r} = 0.$$
From here it is clear that $n = 15$, just expand the sum and see the symmetry:
\begin{align}
& +15\binom{15}{0}+13\binom{15}{1}+11\binom{15}{2}+9\binom{15}{3}+\dots+\binom{15}{7} + \\
&-15\binom{15}{15}-13\binom{15}{14}-11\binom{15}{13}-9\binom{15}{12}-\dots-\binom{15}{8}.
\end{align}
A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{r=0}^n(2r+1)\binom{n}{r}}
&=2\sum_{r=1}^n\binom{n}{r}r+\sum_{r=0}^n\binom{n}{r}\tag{1}\\
&=2n\sum_{r=1}^n\binom{n-1}{r-1}+2^n\tag{2}\\
&=2n\sum_{r=0}^{n-1}\binom{n-1}{r}+2^n\tag{3}\\
&=2n\cdot 2^{n-1}+2^n\tag{4}\\
&\color{blue}{=(n+1)2^n}
\end{align*}

Comment:


*

*In (1) we multiply out. We also start the lower limit of the left-hand sum with $r=1$ which is admissible since we have the factor $0$ in case $r=0$.

*In (2) we apply the binomial identity $\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}$. We also apply the binomial theorem in the form
$\sum_{r=0}^{n}\binom{n}{r}=\sum_{r=0}^{n}\binom{n}{r}1^r1^{n-r}=(1+1)^n=2^n$.

*In (3) we shift the index to start with $r=0$.

*In (4) we again apply the binomial theorem in the form $(1+1)^{n-1}=2^{n-1}$.
A: Just an alternative approach:
$$ \sum_{r=0}^{n}(2r+1)\binom{n}{r}=\frac{d}{dx}\left.\sum_{r=0}^{n}\binom{n}{r}x^{2r+1}\right|_{x=1}=\frac{d}{dx}\left.x(1+x^2)^n\right|_{x=1}=2^n(n+1)$$
hence if this number equals $2^n\cdot 16$ we must have $n=\color{red}{15}$.
A: $$S=\sum_{r=0}^{n}(2r+1)\binom{n}{r}$$
Reversing the order of summation(summing backwards) we have:
$$S=\sum_{r=0}^{n}(2(n-r)+1)\binom{n}{n-r}$$
Since $$\binom{n}{r}=\binom{n}{n-r}$$
We have$$S=\sum_{r=0}^{n}(2(n-r)+1)\binom{n}{r}$$
Now $$2S= \sum_{r=0}^{n}(2r+1)\binom{n}{r}+ \sum_{r=0}^{n}(2(n-r)+1)\binom{n}{r}$$
$$=\sum_{r=0}^{n}(2n+2)\binom{n}{r}$$
Hence $$S= \sum_{r=0}^{n}(n+1)\binom{n}{r}=(n+1) 2^{n}$$
Now Let $(n+1)2^n = 2^{n+4}$, then $n+1= 16$, hence $n=15$
