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So I was messing around, and just thought of a random sequence: $$a_1=3,a_2=5,a_n=a_{n-1}a_{n-2}$$ I wanted to then find the sum of the first $n$ terms of this, and started as so: $$\sum_{k=1}^n a_n = 3+5+3\dot{}5+3\dot{}5^2+3^2\dot{}5^3+... = 8+\sum_{k=2}^{n-1}3^{F(k-1)}\dot{}5^{F(k)}$$where F(k) is the kth Fibbonacci Number. And from here I felt I had reached a dead end. By messing around with the Fibbonacci sequence, I found $$F(n)=F(m)F(n-m-1)+F(m-1)F(n-m) \forall m\in[0,n]$$ When substituting this into my new sum, It makes it intensely complicated regardless of the $m$ chosen. I'm not sure where to go from here, or if I've gone anywhere at all. Better approaches, anything I'm not seeing?

Vedvart1
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    Instead, we may consider sequences of the form $a_n=\exp(A\alpha^n+B\beta^n)$, where $A,B,\alpha,\beta$ are constants. Yet I don't know how to simplify $\sum_{k=1}^n\exp(A\alpha^k+B\beta^k)$. – Yuxiao Xie Apr 21 '17 at 14:20
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    @YuxiaoXie. Such expressions generally do not simplify. – DanielWainfleet Apr 21 '17 at 14:52
  • @user254665 What does it mean by "generally"? – Yuxiao Xie Apr 24 '17 at 04:57
  • I'm sure there are cases where it will simplify, but it usually won't if you encounter it naturally. My running question is whether we could simplify the whole sum at all, not necessarily to a straight out algebraic expression. – Vedvart1 Apr 24 '17 at 05:24

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