The discussion focuses on the recursive sequences defined by the equations $a_{n+1}=5a_n-2b_n$ and $b_{n+1}=a_n+2b_n$, with initial conditions $a_1=1$ and $b_1=-1$. The values of $a_n$ and $b_n$ can be derived through iterative calculation or by solving the recurrence relations. The explicit formulas for $a_n$ and $b_n$ are established as $a_n = 3 \cdot 2^n - 2$ and $b_n = 2^n - 1$. These results provide a clear method for determining the terms of the sequences for any integer n.
PREREQUISITESMathematicians, students studying discrete mathematics, and anyone interested in solving recursive sequences and understanding their applications in various fields.
Albert said:$a_{n+1}=5a_n-2 b_n$
$b_{n+1}=a_n+2 b_n$
$given :\,\, a_1=1,b_1=-1$
$find :$
$a_n,\,\,and,\,\, b_n$
excellent(Yes)Pranav said: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 (***),
$$b_n=6a_{n-1}-a_n=6(4^{n-1}-3^{n-1})-(4^n-3^n)=4^{n-1}\cdot 2 -3^{n-1}\cdot 3$$
$$\Rightarrow b_n=2\cdot 4^{n-1}-3^n$$
$\blacksquare$