Let $a_0 = -2, b_0 = 1,$ and for $n \geq 0,$ let $$a_{n+1} = a_n + b_n + \sqrt{a_n^2 + b_n^2}$$ $$b_{n+1} = a_n + b_n - \sqrt{a_n^2 + b_n^2}.$$ Find $\frac{1}{a_{2012}} + \frac{1}{b_{2012}}.$
Originally I tried looking for a pattern for smaller $n,$ but I couldn't identify anything that was useful to me. Can someone give me a hint as to where I should start? Thanks.
On
Perhaps a bit more roundabout than cr001's solution, but I think it's worth elaborating on and perhaps describing the puzzle-solving process here briefly if that relation amongst the reciprocal sums isn't immediately apparent.
Hint:
First, note that
$$\frac{1}{a_n} + \frac{1}{b_n} = \frac{a_n + b_n}{a_n b_n}$$
We simply want the case where $n=2012$, but sometimes these sort of generalizations help.
So we want some relations by which we can get sums and products of these easily. You can achieve these relations if you add your recurrence relations and multiply them.
Solution (Additive Relation):
Well, if you add the pair together, you see that
$$a_{n+1} + b_{n+1} = 2(a_n+b_n)$$
If we define $c_n := a_n + b_n$, this becomes $c_{n+1} = 2c_n$. This obviously has the solution $c_n = 2^n c_0$. Since $c_0 = a_0 + b_0 = -1$, then $c_n = -2^n$. Thus,
$$a_n + b_n = -2^n$$
Solution (Multiplicative Relation):
Let's try, for the sake of looking at things, multiplying $a_{n+1}$ and $b_{n+1}$. Then, after some simplification, we see
$$a_{n+1} b_{n+1} = 2 a_n b_n$$
So, similar to before, if we let $d_n := a_n b_n$, then $d_n = 2^n d_0$. Then $d_0 = a_0 b_0 = -2$, so
$$d_n = a_n b_n = -2^{n+1}$$
Solution (Final Result):
At this point, we see by invoking our initial formula and the previous two,
$$\frac{1}{a_n} + \frac{1}{b_n} = \frac{-1 \cdot 2^n}{-1 \cdot 2^{n+1}} = \frac 1 2$$
totally independent of the value of $n$! Admittedly this is a curious result, but it seems to hold up under checking the first few values in Excel:
$${1\over a_{n+1}}+{1\over b_{n+1}}={2a_n+2b_n\over (a_n+b_n)^2-{a_n}^2-{b_n}^2}={a_n+b_n\over a_nb_n}={1\over a_n}+{1\over b_n}$$