# Solution for a recurrence relation [closed]

I want to find the solution of the recurrence relation $$a_n =a_{n–1} +2a_{n–2} \text{ with } a_0 =2 \text{ and } a_1 =7$$

thanks for any help.

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what have you tried? –  Prahlad Vaidyanathan Sep 24 at 8:12
maybe $a_n=a_{n-1}+2a_{n-2}$ –  Adi Dani Sep 24 at 8:19
@experimentX, Adi Dani: ya its a_n =a_{n–1} +2a_{n–2} –  steve Sep 24 at 8:21
@steve know of characteristics equations? –  experimentX Sep 24 at 8:22
@steve: Look at my answer here: math.stackexchange.com/questions/154667/… –  Dennis Gulko Sep 24 at 8:59
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## closed as off-topic by azimut, mau, Matt Pressland, Peter Taylor, Daniel RustSep 24 at 9:42

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If the recurrence equation is: $$a_n=a_{n-1}+2a_{n-2}$$ the solution is: $$a_n=c_1(-1)^n+c_22^n$$ So, with your initial conditions you have: $c_1=-1,c_2=3$
$a_n=a\lambda_1^n+b\lambda_2^n$, where $\lambda_1,\lambda_2$ are solutions of equation $\lambda^2=\lambda+2$. From the relations for $a_0$, $a_1$ we can find constants $a,b$: $$a_0=2=a+b,$$ $$a_1=7=2a-b,$$ from where is $a_n=3*2^n-(-1)^n$.