ISI question

Post solution.

a_n+ b_n = a_n+1

also b_n= 2a_n+ b_n or b_n+1 - a_n+1= a_n___(1)

or b_n- a_n= a_n-1 (n≥2)________(2)

From (1) and (2) a_n+2= 2a_n+1 + a_n (n≥1)

Again, a_n+1= a_n+ b_n and (b_n+1- b_n)/2= a_n __(3)

which gives 2a_n+1= b_n+ b_n+1____(4)

We can also write (3) as (b_n+2 - b_n+1)/2= a_n+1 , Combining this with (4) gives

b_n+2= 2b_n+1+ b_n (n≥1)

Now, From the Bold equations it is possible to find a_n and b_n in a closed form.
This kind of equations are known as difference equations.(If you don't know how to solve them try using google)

You will get

a_n= (1+√2)ⁿ+ (1-√2)ⁿ2√2

and b_n= (1+√2)ⁿ+ (1-√2)ⁿ2

Now you can easily check that lim_{n→∞}b_na_n = √2

and b_n² = 2a_n²+1 if n is even and b_n² = 2a_n²-1 is n is odd.

You may see this for 'how to solve difference equations'.
http://www.targetiit.com/iit-jee-forum/posts/fibonacci-series-630.html