- #1
Albert1
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$a_{n+1}=5a_n-2 b_n$
$b_{n+1}=a_n+2 b_n$
$given :\,\, a_1=1,b_1=-1$
$find :$
$a_n,\,\,and,\,\, b_n$
$b_{n+1}=a_n+2 b_n$
$given :\,\, a_1=1,b_1=-1$
$find :$
$a_n,\,\,and,\,\, b_n$
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 (***),
$$b_n=6a_{n-1}-a_n=6(4^{n-1}-3^{n-1})-(4^n-3^n)=4^{n-1}\cdot 2 -3^{n-1}\cdot 3$$
$$\Rightarrow b_n=2\cdot 4^{n-1}-3^n$$
$\blacksquare$
To find a_n and b_n, we can use the general formula for an arithmetic sequence: a_n = a_1 + (n-1)d and b_n = b_1 + (n-1)d. Here, d represents the common difference between consecutive terms in the sequence. Simply plug in the values of a_1, b_1, and n into the formula to find the values of a_n and b_n.
Yes, you can use any positive integer value for n when finding a_n and b_n. However, keep in mind that the values of a_n and b_n will vary depending on the value of n that you choose.
No, in order to find a_n and b_n, you need to have at least two given terms. This is because you need to know the common difference between terms in order to use the formula mentioned in the first question.
No, this method only applies to arithmetic sequences, where each term is obtained by adding a fixed number to the previous term. For other types of sequences, different methods may need to be used to find a_n and b_n.
Yes, if the sequence is given in a table or graph, you can visually determine the common difference between terms and use that to find a_n and b_n. This can be a faster method than using the general formula.