How find the $a_{10}+a_{2014}$ if $a_{n+1}=\frac{8}{5}a_{n}+\frac{6}{5}\sqrt{4^n-a^2_{n}}$ 
The sequence $\{a_{n}\}$ satisfies $a_{0}=1$,and
  $$a_{n+1}=\dfrac{8}{5}a_{n}+\dfrac{6}{5}\sqrt{4^n-a^2_{n}},n\ge 0$$
Find the $a_{10}+a_{2014}$.

My idea: since
$$5a_{n+1}-8a_{n}=6\sqrt{4^n-a^2_{n}},n\ge 0$$
so
$$25a^2_{n+1}-80a_{n+1}a_{n}+64a^2_{n}=36(4^n-a^2_{n})$$
$$\Longrightarrow 25a^2_{n+1}-80a_{n+1}a_{n}+100a^2_{n}=36\cdot 4^n\tag{1}$$
and
$$25a^2_{n}-80a_{n}a_{n-1}+100a^2_{n-1}=36\cdot 4^{n-1}$$
$$\Longrightarrow 100a^2_{n}-320a_{n}a_{n-1}+400a^2_{n-1}=36\cdot 4^n\tag{2}$$
then $(1)-(2)$,we have
$$25(a^2_{n+1}-16a^2_{n-1})=80a_{n}(a_{n+1}-4a_{n-1})$$
so
$$(a_{n+1}-4a_{n-1})(5a_{n+1}-16a_{n}+20a_{n-1})=0$$
so
$a_{n+1}=4a_{n-1}$,
or
$5a_{n+1}=16a_{n}-20a_{n-1}$
and I find this ugly,But I fell very ugly.maybe have other methods,it is said can find the $a_{n}$ close form.
$$a_{n}=2^n\sin{x_{n}}?$$.
But I can't.Thank you very much

We have $a_0=1, a_1=\frac{8}{5}, a_2=4, a_3=\frac{32}{5}$. This leads us to see the pattern $$a_{2n}=2^{2n}, a_{2n+1}=\frac{2^{2n+3}}{5}$$ This is easily proven by induction. Thus $a_{10}+a_{2014}=2^{10}+2^{2014}$.
