Sum $\sum_{k=0}^{2013}2^ka_{k}$ let real sequence $a_{0},a_{1},a_{2},\cdots,a_{n}$,such
$$a_{0}=2013,a_{n}=-\dfrac{2013}{n}\sum_{k=0}^{n-1}a_{k},n\ge 1$$
How find this sum 
$$\sum_{k=0}^{2013}2^ka_{k}$$
My idea: since
$$-na_{n}=2013(a_{0}+a_{1}+a_{2}+\cdots+a_{n-1})\cdots\cdots(1)$$
so
$$-(n+1)a_{n+1}=2013(a_{0}+a_{1}+\cdots+a_{n})\cdots\cdots (2)$$
then
$(2)-(1)$,we have
$$na_{n}-(n+1)a_{n+1}=2013a_{n}$$
then
$$(n+1)a_{n+1}=(2013-n)a_{n}$$
then 
$$\dfrac{a_{n+1}}{a_{n}}=\dfrac{2013-n}{n+1}$$
so
$$\dfrac{a_{n}}{a_{n-1}}\cdot\dfrac{a_{n-1}}{a_{n-2}}\cdots\dfrac{a_{1}}{a_{0}}=\cdots$$
so
$$\dfrac{a_{2013}}{a_{0}}=\dfrac{1}{2013}\cdot\dfrac{2}{2012}\cdots\dfrac{2013}{1}=1?$$
then How can find this sum?
 A: Method 1 (generating function)
Let $a(z) = \sum\limits_{k=0}^\infty a_k z^k$, notice
$$a(z)\left(\frac{z}{1-z}\right) = \left(\sum_{k=0}^\infty a_k\right)\left( \sum_{\ell=1}^\infty z^\ell\right) = \sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}a_k\right)z^n$$
and 
$$\left(z\frac{d}{dz}\right)a(z) = \left(z\frac{d}{dz}\right)\sum_{n=0}^\infty a_nz^n
= \sum_{n=1}^\infty na_n z^n$$
The equality $a_{n}=-\dfrac{2013}{n}\sum_{k=0}^{n-1}a_{k},\,n\ge 1$ implies
$$\frac{da(z)}{dz} = -2013\frac{a(z)}{1-z}
\quad\iff\quad \frac{d}{dz} \log a(z) = 2013\frac{d}{dz}\log(1-z)$$
and hence
$$a(z) = a_0(1-z)^{2013} = 2013(1-z)^{2013}$$
Since this is a polynomial with degree 2013, we get
$$\sum_{k=0}^{2013} a_k 2^k = a(2) = 2013 (1-2)^{2013} = -2013$$
Method 2 (more elementary, appropriate for middle school students)
Let $b_n = \sum\limits_{k=0}^n a_k$, we have $b_0 = 2013$ and for $n > 0$,
$$n(b_n-b_{n-1}) = -2013 b_{n-1}\quad\iff\quad  b_n = -\frac{2013 -n}{n}b_{n-1}$$
This implies
$$b_n = (-1)^n \frac{\prod\limits_{k=1}^n (2013-k)}{n!} b_0
= (-1)^n \frac{2013!}{n!(2013-n-1)!}
= (-1)^n \binom{2013}{n} (2013-n)$$
Notice $$b_{n-1} = (-1)^{n-1}\frac{2013!}{(n-1)!(2013-n)!} = (-)^{n-1} \binom{2013}{n} n,$$
we obtain
$$a_n = b_n - b_{n-1} = (-1)^n 2013\binom{2013}{n}$$
Using binomial theorem, we can evaluate the desired sum as
$$\sum_{k=0}^{2013} a_k 2^k = 2013 \sum_{k=0}^{2013} \binom{2013}{k}(-2)^k = 2013 (1-2)^{2013} = -2013$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{a_{0}=2013\,,\quad a_{n} = -\,{2013 \over n}\sum_{k = 0}^{n - 1}a_{k}\,,\quad
     n \geq 1:\ {\large ?}}$

\begin{align}
&\pars{n + 1}a_{n + 1} - na_{n}=-2013\pars{\sum_{k = 0}^{n}a_{k} - \sum_{k = 0}^{n - 1}a_{k}} = -2013 a_{n}
\\[3mm]&\imp\quad
\pars{n + 1}a_{n + 1} - \pars{n - a_{0}}a_{n} =  0
\end{align}

\begin{align}
&\sum_{n = 0}^{\infty}\pars{n + 1}a_{n + 1}z^{n} - \sum_{n = 0}^{\infty}na_{n}z^{n}
+a_{0}\sum_{n = 0}^{\infty}a_{n}z^{n}=0
\\[3mm]&
\sum_{n = 1}^{\infty}na_{n}z^{n - 1}
-z\partiald{}{z}\sum_{n = 0}^{\infty}a_{n}z^{n}
+a_{0}\sum_{n = 0}^{\infty}a_{n}z^{n}=0
\\[3mm]&
\partiald{}{z}\sum_{n = 0}^{\infty}a_{n}z^{n}
-z\partiald{}{z}\sum_{n = 0}^{\infty}a_{n}z^{n}
+a_{0}\sum_{n = 0}^{\infty}a_{n}z^{n}=0
\\[3mm]&
\partiald{}{z}\sum_{n = 0}^{\infty}a_{n}z^{n}
-{a_{0} \over z - 1}\sum_{n = 0}^{\infty}a_{n}z^{n}=0
\\[3mm]&
\pars{z - 1}^{-a_{0}}\partiald{}{z}\sum_{n = 0}^{\infty}a_{n}z^{n}
-a_{0}\pars{z - 1}^{-a_{0} - 1}\sum_{n = 0}^{\infty}a_{n}z^{n}=0
\\[3mm]&
\partiald{}{z}
\bracks{\pars{z - 1}^{-a_{0}}\sum_{n = 0}^{\infty}a_{n}z^{n}}=0
\quad\imp\quad\pars{z - 1}^{-a_{0}}\sum_{n = 0}^{\infty}a_{n}z^{n} = -a_{0}
\end{align}

\begin{align}
&\sum_{n = 0}^{\infty}a_{n}z^{n}=a_{0}\pars{1 - z}^{a_{0}}
=a_{0}\sum_{n = 0}^{a_{0}}{a_{0} \choose n}\pars{-1}^{n}z^{n}
\end{align}

With $\ds{z = -2}$:
$$
\color{#00F}{\large\sum_{k = 0}^{2013}2^{k}a_{k}} = 2013\pars{1 - 2}^{2013} = \color{#00F}{\large -2013}
$$
A: oh sorry, i omit some details !
by the symmetric property, $a_{0}=-a_{2013}$$,$$a_{1}=-a_{2012}$$,......$
the hint is :
$(a_{2}+4/3a_{2}+......+4/3a_{3}+......)/(ta_{0})=-a_{1}/a_{0}(a_{1}-4/3a_{1}-......-4/3a_{2}-......)$
we assume that :
$t_{1}=\frac{(a_{2}+4/3a_{2}+......+4/3a_{3}+......)}{(-a_{1}-4/3a_{1}-......-4/3a_{2}-......)}$
$t_{2}=\frac{(a_{3}+4/3a_{3}+......+4/3a_{4}+......)}{(-a_{2}-4/3a_{2}-......-4/3a_{3}-......)}$
so,
$a_{0}(1+1+4/3+......-a_{0}-4/3a_{0}-......-4/3\cdot1/2(a_{0}+a_{1})-......)$$=$$(\frac{a_{0}}{a_{1}}\cdot{\frac{a_{1}}{a_{2}}}$$\cdot......\cdot{\frac{a_{2012}}{a_{2013}}})\cdot{\frac{a_{0}}{a_{2013}}}\cdot$$t_{1}\cdot{t_{2}}......\cdot{t_{n}}$$=t_{1}\cdot{t_{2}}......\cdot{t_{n}}$
$=\frac{(a_{n-1}+4/3a_{n-1}+......+4/3a_{n-2}+......)}{(-a_{1}-4/3a_{1}-......-4/3a_{2}-......)}=1$
therefore, your solution holds !
