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Show that the sequence {an} is a solution of the recurrence relation an = -3an-1 + 4an-2 if

a) an = 0

b) an = 1

c) an = (-4)n

d) an = 2(-4)n + 3

In the question What is sequence {an} ??

And how to solve this kind of question?

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$a)a_{n}=0$

$R.H.S=-3a_{n-1}+4a_{n-2}=-3\times 0+4\times 0=0$ [because 0 is a constant value]

$b)a_{n}=1$

$R.H.S=-3a_{n-1}+4a_{n-2}=-3\times 1+4\times 1=1$

$c)a_{n}=(-4)^{n}$

$R.H.S=-3a_{n-1}+4a_{n-2}=-3\times (-4)^{n-1}+4\times \left ( -4 \right )^{n-2}$

$=12(-4)^{n-2}+4 \left ( -4 \right )^{n-2}$

$=16\left ( -4 \right )^{n-2}=(-4)^{n}$

$d)a_{n}=2(-4)^{n}+3$

$R.H.S=-3\left [ 2\left ( -4 \right )^{n-1} +3\right ]+4\left [ 2\left ( -4 \right )^{n-2} +3\right ]$

$=24\left ( -4 \right )^{n-2} -9+8\left ( -4 \right )^{n-2} +12$

$=32\left ( -4 \right )^{n-2}+3$

$=2\left ( -4 \right )^{n}+3$

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