Calculation of $\frac{a_{20}}{a_{20}+b_{20}}$?

Question

The solution of $\frac{a_{20}}{a_{20}+b_{20}}$ is $-39$ (This is wrote by answer sheet) from the recursive system of equations :

\begin{cases} a_{n+1}=-2a_n-4b_n \\ b_{n+1}=4a_n+6b_n\\ a_0=1,b_0=0 \end{cases}

This is taken from $2007$ GATE entrance exam in India.

anyone can show me how we can calculate this answer?

Update 1:

Three answer is added, but my main problem is remains up to yet, non of these three answers didn't include the main aspect of this question. my main problem is via simplification and replacement in last part of solution.

Answer 1

Hint :

show that

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

Answer 2

$4b_n=-a_{n+1}-2a_n$, $4b_{n+1}=-a_{n+2}-2a_{n+1}$, $-a_{n+2}-2a_{n+1}=16a_n-6a_{n+1}-12a_n$, $a_{n+2}-4a_{n+1}+4a_n=0$. Do you know how to solve that kind of recurrence?

Here's an approach. $a_{n+2}-4a_{n+1}+4a_n=(a_{n+2}-2a_{n+1})-2(a_{n+1}-2a_n)=c_{n+1}-2c_n$, where we are defining $c_n$ by $c_n=a_{n+1}-2a_n$. Now we have to solve $c_{n+1}-2c_n=0$, and the solution is obviously $c_n=c_02^n$ (and we can work out $c_0$ easily enough). So now we have to solve $a_{n+1}-2a_n=c_02^n$. Are we at a recurrence you can solve yet?

Answer 3

$$a_{n+1}=-2a_n-4b_n \tag1$$ $$b_{n+1}=4a_n+6b_n \tag2$$ $$a_0=1,b_0=1 \tag3$$

$$(1) + (2) \implies a_{n+1}+b_{n+1}=2a_n+2b_n=2(a_n+b_n) $$ $$\implies \frac{a_{n+1}+b_{n+1}}{a_n+b_n}=2$$

Similarly we have $$\frac{a_{n}+b_{n}}{a_{n-1}+b_{n-1}}=\frac{a_{n-1}+b_{n-1}}{a_{n-2}+b_{n-2}}=\ldots=\frac{a_1+b_1}{a_0+b_0}=2$$

Therefore we can write that $$\frac{a_{n+1}+b_{n+1}}{a_n+b_n}\cdot\frac{a_{n}+b_{n}}{a_{n-1}+b_{n-1}}\cdot\frac{a_{n-1}+b_{n-1}}{a_{n-2}+b_{n-2}} \ldots \frac{a_1+b_1}{a_0+b_0}=2\cdot2\cdot2\ldots2$$ $$\implies \frac{a_{n+1}+b_{n+1}}{a_0+b_0}=2^{n+1}$$ $$\implies \frac{a_{n+1}+b_{n+1}}{1+1}=2^{n+1}$$ $$\implies a_{n+1}+b_{n+1}=2^{n+2}$$

So, $$a_{20}+b_{20}=2^{21}$$

Now, you can find $a_n$ from the relation $a_{n+2}-4a_{n+1}+4a_n=0$ by using the power series method. And using the above relation, you can deduce $b_n$.

Hope this helps.