# recurrence relation

• April 15th 2009, 10:55 AM
othnin
recurrence relation
So I understand how to solve it the questions but I think my knowledge of exponents is causing me problems:

Determine if the sequence {a(n)} is a solution of the recurrence relation a(n) = -3*a(n-1) + 4*a(n-2) if a(n) = 2*(-4)^n+3

I started as below substituting in the formula:

a(n) = -3( 2(-4)^(n-1)+3) + 4( 2(-4)^(n-2) + 3)
= ( -3 *2* -4^(n-1) -9) + 2 * -4^(n-1) + 12

However,
I am having trouble with the first term. Any help would be appreciated.
• April 15th 2009, 05:45 PM
xalk
Quote:

Originally Posted by othnin
So I understand how to solve it the questions but I think my knowledge of exponents is causing me problems:

Determine if the sequence {a(n)} is a solution of the recurrence relation a(n) = -3*a(n-1) + 4*a(n-2) if a(n) = 2*(-4)^n+3

I started as below substituting in the formula:

a(n) = -3( 2(-4)^(n-1)+3) + 4( 2(-4)^(n-2) + 3)
= ( -3 *2* -4^(n-1) -9) + 2 * -4^(n-1) + 12

However,
I am having trouble with the first term. Any help would be appreciated.

$a_{n} =2(-4)^3(-4)^n =-128(-4)^n$

$a_{n-1} = 2(-4)^2(-4)^n = 32(-4)^n$

$a_{n-2} = 2(-4)(-4)^n = -8(-4)^n$

hence $a_{n} =-3a_{n-1} + 4a_{n-2}$
• April 15th 2009, 06:20 PM
Soroban
Hello, othnin!

You started off fine . . .

Quote:

Determine if $a(n) \:=\:2\!\cdot\!(\text{-}4)^{n+3}$ is a solution of the recurrence relation:

. . $a(n) \:=\:-3\!\cdot\!a(n-1) + 4\!\cdot\!a(n-2)$

Substitute: . $a(n) \;=\;-3\bigg[2(\text{-}4)^{n+2}\bigg] + 4\bigg[2(\text{-}4)^{n+1}\bigg]$

. . . . . . . . . . . . $= \;-6(\text{-}4)^{n+1} + 8(\text{-}4)^{n+1}$

. . . . . . . . . . . . $= \;2\cdot(\text{-}4)^{n+1}\cdot\bigg[-3(\text{-}4) + 4\bigg]$

. . . . . . . . . . . . $= \;2\cdot(\text{-}4)^{n+1}\cdot\bigg[12 + 4\bigg]$

. . . . . . . . . . . . $= \;2\cdot(\text{-}4)^{n+1}\cdot16$

. . . . . . . . . . . . $= \;2\cdot(\text{-}4)^{n+1}\cdot(\text{-}4)^2$

. . . . . . . . . . . . $= \;2\cdot(\text{-}4)^{n+3}\qquad\hdots$ It's true!