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I am having trouble finding initial conditions for closed sequences. Could you help me get through this problem from my textbook? Thanks!

Show that $3 · 2^n + 7 · 5^n$ is a solution to the recurrence relation, $a_{n} = 7a_{n−1} − 10a_{n−2}$. What would the initial conditions need to be for this to be the closed formula for the sequence?

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  • $\begingroup$ Can you not simply plug in $n = 0, 1, 2$ into the expression? $\endgroup$ – Brian Tung Mar 22 '18 at 19:43
  • $\begingroup$ Yes thank you @BrianTung $\endgroup$ – Ethan Walser Mar 22 '18 at 19:48
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The initial conditions are $a_0=3+7=10$ and $a_1=3\times2+7\times5=41$, of course.

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  • $\begingroup$ That means $a_n = 7a_{n−1} − 10a_{n−2}$? $\endgroup$ – Mathew Mahindaratne Mar 22 '18 at 19:47
  • $\begingroup$ @MathewMahindaratne The only sequence that satisfies the initial conditions that I mentioned and such that $a_n=7a_{n-1}-10a_{n-2}$ is the one mentioned by the OP. $\endgroup$ – José Carlos Santos Mar 22 '18 at 19:49
  • $\begingroup$ Thanks for clarification. $\endgroup$ – Mathew Mahindaratne Mar 22 '18 at 19:57

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