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find a_n and b_n

Albert

Well-known member
Jan 25, 2013
1,225
$a_{n+1}=5a_n-2 b_n$
$b_{n+1}=a_n+2 b_n$
$given :\,\, a_1=1,b_1=-1$
$find :$
$a_n,\,\,and,\,\, b_n$
 

Pranav

Well-known member
Nov 4, 2013
428
$a_{n+1}=5a_n-2 b_n$
$b_{n+1}=a_n+2 b_n$
$given :\,\, a_1=1,b_1=-1$
$find :$
$a_n,\,\,and,\,\, b_n$

We have:
$$a_{n+1}=5a_n-2b_n\,\,\,\, (*)$$
$$b_{n+1}=a_n+2b_n\,\,\,\, (**)$$
Add (*) and (**) to get:
$$b_{n+1}+a_{n+1}=6a_n \Rightarrow b_{n+1}=6a_n-a_{n+1} \Rightarrow b_n=6a_{n-1}-a_n\,\,\, (***)$$
Substitute (***) in (*) to get:
$$a_{n+1}=5a_n-12a_{n-1}+2a_n \Rightarrow a_{n+1}=7a_n-12a_{n-1}$$
The solution of above recursive relation is of the form $a_n=c_1 4^n+c_2 3^n$. We are given $a_1=1$. $a_2$ can be found from (*) and comes out to be 7. From these two conditions, $c_1=1$ and $c_2=-1$. Hence $a_n=4^n-3^n$.

From (***),
$$b_n=6a_{n-1}-a_n=6(4^{n-1}-3^{n-1})-(4^n-3^n)=4^{n-1}\cdot 2 -3^{n-1}\cdot 3$$
$$\Rightarrow b_n=2\cdot 4^{n-1}-3^n$$
$\blacksquare$
 

Albert

Well-known member
Jan 25, 2013
1,225

We have:
$$a_{n+1}=5a_n-2b_n\,\,\,\, (*)$$
$$b_{n+1}=a_n+2b_n\,\,\,\, (**)$$
Add (*) and (**) to get:
$$b_{n+1}+a_{n+1}=6a_n \Rightarrow b_{n+1}=6a_n-a_{n+1} \Rightarrow b_n=6a_{n-1}-a_n\,\,\, (***)$$
Substitute (***) in (*) to get:
$$a_{n+1}=5a_n-12a_{n-1}+2a_n \Rightarrow a_{n+1}=7a_n-12a_{n-1}$$
The solution of above recursive relation is of the form $a_n=c_1 4^n+c_2 3^n$. We are given $a_1=1$. $a_2$ can be found from (*) and comes out to be 7. From these two conditions, $c_1=1$ and $c_2=-1$. Hence $a_n=4^n-3^n$.

From (***),
$$b_n=6a_{n-1}-a_n=6(4^{n-1}-3^{n-1})-(4^n-3^n)=4^{n-1}\cdot 2 -3^{n-1}\cdot 3$$
$$\Rightarrow b_n=2\cdot 4^{n-1}-3^n$$
$\blacksquare$
excellent(Yes)