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fiksx

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Post moved by moderator, so missing the homework template.

series ##{a_n}## is define by ##a_1=1 ## , ##a_2=5 ## , ##a_3=1 ##, ##a_{n+3}=a_{n+2}+4a_{n+1}-4a_n ## ( ##n \geq 1 ##).

$$\begin{pmatrix}a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ \end{pmatrix}=B\begin{pmatrix}a_{n+2} \\ a_{n+1} \\ a_{n} \\ \end{pmatrix}$$

find ##B##

find general term of ##a_n## series

I found matrix b is ##B=\begin{bmatrix}1&4&-4\\1&0&0\\0&1&0\end{bmatrix} ##

but how can I find ##a_n ##?

I can compute that ##\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}a_{3} \\ a_{2} \\ a_{1} \\ \end{pmatrix} ##

##\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} ##

##\begin{pmatrix}a_{5} \\ a_{4} \\ a_{3} \\ \end{pmatrix}=B^2\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} ##

can I just plug In ##n=1## to the equation?

i see there is relation about this.

##a_{n+1}=B^na_n ##<br> suppose D is diagonal matrix that similar to B <br>

##a_{n+1}=TD^nT^- a_n ##

which mean i need to find diagonal matrix and its inver to find ##a_n ##? is my assumption wrong?

but the right formula is ##\begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_n \end{pmatrix}=B^{n-1}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix}## , how can I get this? and to find ##a_n## should I find ##B^n## or ##B^{n-1}##?

series ##{a_n}## is define by ##a_1=1 ## , ##a_2=5 ## , ##a_3=1 ##, ##a_{n+3}=a_{n+2}+4a_{n+1}-4a_n ## ( ##n \geq 1 ##).

$$\begin{pmatrix}a_{n+3} \\ a_{n+2} \\ a_{n+1} \\ \end{pmatrix}=B\begin{pmatrix}a_{n+2} \\ a_{n+1} \\ a_{n} \\ \end{pmatrix}$$

find ##B##

find general term of ##a_n## series

I found matrix b is ##B=\begin{bmatrix}1&4&-4\\1&0&0\\0&1&0\end{bmatrix} ##

but how can I find ##a_n ##?

I can compute that ##\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}a_{3} \\ a_{2} \\ a_{1} \\ \end{pmatrix} ##

##\begin{pmatrix}a_{4} \\ a_{3} \\ a_{2} \\ \end{pmatrix}=B\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} ##

##\begin{pmatrix}a_{5} \\ a_{4} \\ a_{3} \\ \end{pmatrix}=B^2\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix} ##

can I just plug In ##n=1## to the equation?

i see there is relation about this.

##a_{n+1}=B^na_n ##<br> suppose D is diagonal matrix that similar to B <br>

##a_{n+1}=TD^nT^- a_n ##

which mean i need to find diagonal matrix and its inver to find ##a_n ##? is my assumption wrong?

but the right formula is ##\begin{pmatrix} a_{n+2} \\ a_{n+1} \\ a_n \end{pmatrix}=B^{n-1}\begin{pmatrix}1 \\ 5 \\ 1 \\ \end{pmatrix}## , how can I get this? and to find ##a_n## should I find ##B^n## or ##B^{n-1}##?

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