Find value of $a_{2012}$

Question

A sequence $\left\{a_n\right\}$ is defined as:

$a_1=1$, $a_2=2$ and

$$a_{n+1}=\frac{2}{a_n}+a_{n-1}$$ $\forall$ $n \ge 2$

Find $a_{2012}$

My Try:

we have

$$a_{n+1}-a_{n-1}=\frac{2}{a_n}$$

$$a_n a_{n+1}-a_{n-1}a_n=2 \tag{1}$$

Replacing $n$ with $n-1$ we get

$$a_{n-1} a_{n}-a_{n-2}a_{n-1}=2 \tag{2}$$ adding $(1)$ and $(2)$

we get

$$a_n a_{n+1}-a_{n-2}a_{n-1}=4 \tag{3}$$

Again replace $n$ with $n-1$ in $(3)$ and adding with $(1)$ we get

$$a_n a_{n+1}-a_{n-2}a_{n-3}=6 \tag{4}$$ Again replace $n$ with $n-1$ in $(4)$ and adding with $(1)$ we get

$$a_n a_{n+1}-a_{n-3}a_{n-4}=8 \tag{5}$$

Continuing the process we get

$$a_na_{n+1}-a_{n-2010}a_{n-2011}=4022$$

Now in above equation put $n=2012$ we get

$$a_{2012}a_{2013}-a_1a_2=4022$$ $\implies$

$$a_{2012}a_{2013}=4024$$

Any further clue?

Answer 1

You already have $$a_n a_{n+1}-a_{n-1}a_n=2$$

So if you set the new sequence $b_k=a_{k+1}a_{k}$ then you get

$$b_k-b_{k-1}=2$$ so,

$b_k$ is an arithmetic sequence with $b_1=a_1a_2=2$ and step $2$. Then

$$b_k=2+2(k-1)=2k$$

then

$$a_{2}a_{1}=2$$ $$4=a_{3}a_{2}$$ $$a_{4}a_{3}=6$$ $$8=a_{5}a_{4}$$ $$...$$ $$2\cdot 2010=a_{2011}a_{2010}$$ $$a_{2012}a_{2011}=2\cdot2011$$

Multiply every equation and get

$$4\cdot8\cdot12\cdot...2020\cdot (a_{2012}\cdot a_1)=2\cdot6\cdot10\cdot...2022$$

$$a_{2012}=\frac{2\cdot6\cdot10\cdot...2022}{4\cdot8\cdot12\cdot...2020}$$

Answer 2

By your work we obtain: $$a_{n+2}a_{n+1}=a_{n+1}{a_n}+2,$$ which gives $$a_{n+1}a_n=2+(n-1)2=2n.$$ Thus, $$a_{n}=\frac{2n-2}{\frac{2n-4}{a_{n-2}}}=\frac{n-1}{n-2}a_{n-2},$$ which for even $n$ gives $$a_n=\frac{(n-1)(n-3)...1}{(n-2)(n-4)...2}a_2=\frac{2(n-1)!!}{(n-2)!!}.$$ Id est, $$a_{2012}=\frac{2\cdot2011!!}{2010!!}.$$

Answer 3

If you put $b_n=a_na_{n+1}$, your relation (1) give that $b_n-b_{n-1}=2$; from this you easily prove that $b_n=2n$ for all $n$, ie $a_na_{n+1}=2n$. Now $a_{n+2}=2(n+1)/a_{n+1}=(1+1/n)a_n$, and if $c_k=a_{2k}$, you have $c_{k+1}=\frac{2k+1}{2k} c_k$; it is easy to finish.