This problem is from a math competition from 1994. I have been having trouble working with this problem:
Let $f(1) = 1, $ and $f(n + 1) = 2\sqrt{f(n)^2 + 1}$ for $n \geq 1$. If $N \geq 1$ is an integer, find $$\sum_{n = 1}^{N}f(n)^2.$$
I tried many things, and although I found some patterns, I couldn't really get anywhere. Some values, which might help, are listed below:
- $f(1) = 1$
- $f(2) = 2\sqrt{2}$
- $f(3) = 6$
- $f(4) = 2\sqrt{37}$
- $f(5) = 2\sqrt{149}$
Since the problem is so old, there is no solution provided.
If we put $b_n = f(n)^2$ then we have $$b_{n+1} -4b_n = 4 = b_n-4b_{n-1}\implies \boxed{b_{n+1} -5b_n +4b_{n-1}=0}$$
and we have to find $$S:=\sum_{n=1}^Nb_n$$
Solving boxed characteristic equation $x^2-5x+4=0$ we get $b_n = a+b\cdot 4^n$. If we take in to a count that $b_1=1$ and $b_2=8$ we get $b_n = {1\over 3}(7\cdot 4^{n-1}-4) $
So $$ S = {1\over 3}\Big(7{4^N-1\over 4-1} -4N\Big) = {1\over 9}\Big(7\cdot 4^N-7-12N\Big)$$