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?

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 !
