4

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

math110
  • 93,304

1 Answers1

2

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 $a_{10}+a_{2014}=2^{10}+2^{2014}$.

Ivan Loh
  • 16,955