Consider the sequence a_1,a_2,a_3, such that a_1=1, a_2=2, and a_n+2=2a_n+1+a_n for n=1,2,…

Question

Consider the sequence $a_1,a_2,a_3,\ldots$ such that $a_1=1$, $a_2=2$, and $a_{n+2}=\dfrac{2}{a_{n+1}}+a_n$ for $n=1,2,3,\ldots$. If
$$\left(a_1+\frac{1}{a_2}\right)\!\left(a_2+\frac{1}{a_3}\right)\cdots\left(a_{31}+\frac{1}{a_{32}}\right) = 2^{\alpha}\,{}^{61}C_{31},$$
then $\alpha$ is equal to:

Accepted Answer

The recurrence gives $a_{n+2}a_{n+1}-a_na_{n+1}=2$, so $a_{n+2}a_{n+1} = 2n+a_1a_2$.

Each factor simplifies: $\dfrac{a_na_{n+1}+1}{a_{n+1}a_{n+2}} = \dfrac{2n-1}{2n}$ (using $a_1a_2=2$), so the product telescopes to $\dfrac{3}{4}\cdot\dfrac{5}{6}\cdots\dfrac{61}{62} = 2^{-60}\,{}^{61}C_{31}$.

So $\alpha=-60$.

Solutions · 1