Show that $2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$ 
If n is a positive integer I need to show that 
$2^n-(n-1)2^{n-2}+\frac{(n-2)(n-3)}{2!}2^{n-4}-...=n+1$

My guess: Somehow I need two equivalent binomial expression whose coefficients I need to compare.But which two binomial expressions? I know not!
P.S:Don't use Sterling Numbers or very high level maths...
 A: You can prove it by induction on $n$. Let
$$f(n)=\sum_{k\ge 0}(-1)^k2^{n-2k}\binom{n-k}k\;.$$
Then for the induction step you have
$$\begin{align*}
f(n+1)&=\sum_{k\ge 0}(-1)^k2^{n+1-2k}\binom{n+1-k}k\\
&=2\sum_{k\ge 0}(-1)^k2^{n-2k}\left(\binom{n-k}k+\binom{n-k}{k-1}\right)\\
&=2f(n)-2\sum_{k\ge 0}(-1)^k2^{n-2-2k}\binom{n-1-k}k\\
&=2f(n)-\sum_{k\ge 0}(-1)^k2^{n-1-2k}\binom{n-1-k}k\\
&=2f(n)-f(n-1)\\
&=2(n+1)-n\\
&=n+2\;.
\end{align*}$$
A: Here is a generating function approach
$$
\begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}2^{n-2k}x^n\tag{2}\\
&=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=k}^\infty\binom{n-k}{k}(2x)^n\tag{3}\\
&=\sum_{k=0}^\infty\left(-\frac14\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^{n+k}\tag{4}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty\binom{n}{k}(2x)^n\tag{5}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^{n-k}\binom{-k-1}{n-k}(2x)^n\tag{6}\\
&=\sum_{k=0}^\infty\left(-\frac x2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^{n+k}\tag{7}\\
&=\sum_{k=0}^\infty\left(-x^2\right)^k\sum_{n=0}^\infty(-1)^n\binom{-k-1}{n}(2x)^n\tag{8}\\
&=\sum_{k=0}^\infty\left(-x^2\right)^k\frac1{(1-2x)^{k+1}}\tag{9}\\
&=\frac1{1-2x}\frac1{1+\frac{x^2}{1-2x}}\tag{10}\\
&=\frac1{(1-x)^2}\tag{11}\\
&=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^k\tag{12}\\
&=\sum_{k=0}^\infty(k+1)x^k\tag{13}\\
\end{align}
$$
Explanation:
$\phantom{0}(2)$: change order of summation
$\phantom{0}(3)$: move $(-1)^k2^{-2k}=\left(-\frac14\right)^k$ out front
$\phantom{0}(4)$: substitute $n\mapsto n+k$
$\phantom{0}(5)$: move $(2x)^k$ out front
$\phantom{0}(6)$: $\binom{n}{k}=\binom{n}{n-k}=(-1)^{n-k}\binom{-k-1}{n-k}$ (see this answer)
$\phantom{0}(7)$: substitute $n\mapsto n+k$
$\phantom{0}(8)$: move $(2x)^k$ out front
$\phantom{0}(9)$: Binomial Theorem
$(10)$: sum of a geometric series
$(11)$: simplification
$(12)$: Binomial Theorem
$(13)$: $(-1)^k\binom{-2}{k}=\binom{k+1}{k}=\binom{k+1}{1}=k+1$
Equating the coefficients of $x^k$, we get $a_n=n+1$.
A: We can develop the Linear Recurrence
$$
\begin{align}
a_n
&=\sum_{k=0}^n(-1)^k\binom{n-k}{k}2^{n-2k}\\
&=\sum_{k=0}^n(-1)^k\left[\binom{n-k-1}{k}+\binom{n-k-1}{k-1}\right]2^{n-2k}\\
&=2\sum_{k=0}^{n-1}(-1)^k\binom{n-k-1}{k}2^{n-2k-1}-\sum_{k=0}^{n-2}(-1)^k\binom{n-k-2}{k}2^{n-2k-2}\\[6pt]
&=2a_{n-1}-a_{n-2}
\end{align}
$$
which has the characteristic equation
$$
x^2-2x+1=0
$$
which has a double root at $x=1$. Thus, the solution has the form
$$
a_n=c_0\cdot1^n+c_1n\cdot1^n
$$
Since $a_0=1$ and $a_1=2$, we get
$$
a_n=n+1
$$
A: I leave the judgement of uniqueness to the reader.
Let use consider the $r+1$th term of $$(2x^a-x^b)^{m-r}$$ which will be
$$\binom{m-r}r2^{m-2r}(-1)^rx^{am+r(b-2a)}$$
WLOG choose $b=2,a=1$
So, $\displaystyle\binom{m-r}r2^{m-2r}(-1)^r$ will be the coefficient of $x^m$ of  in $\displaystyle(2x-x^2)^{m-r}$
$$\implies\sum_{r=0}^{2r\le m}\binom{m-r}r2^{m-2r}(-1)^r$$ will be the coefficient of $x^m$ in the expansion of $$\sum_{r=0}^{2r\le m}(2x-x^2)^{m-r}$$
i.e., in the expansion of $$\sum_{r=0}^m(2x-x^2)^{m-r}=\sum_{u=0}^m(2x-x^2)^u=\dfrac{1-(2x-x^2)^{m+1}}{1-(2x-x^2)}=\{1-(2x-x^2)^{m+1}\}(1-x)^{-2}$$
i.e., in the expansion of $\displaystyle(1-x)^{-2}$
Now the coefficient in $x^n(n\ge0)$ in $\displaystyle(1-x)^{-2}$ (assuming the convergence) is
$$\dfrac{(-1)^n(-2)(-3)\cdots(-n)(-n-1)}{n!}=n+1$$
