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- Thread starter goohu
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- Feb 7, 2012

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One way to do this would be to replace $a_k$ by $b_k = a_k + c(-1)^k$ (where $c$ is a constant to be chosen later). Then $a_k = b_k - c(-1)^k$, and the recurrence equation for $a_k$ becomes $$b_k - c(-1)^k = 3(b_{k-1} - c(-1)^{k-1}) - (b_{k-2} - c(-1)^{k-2}) - 2(-1)^k,$$ $$b_k = 3b_{k-1} - b_{k-2} + (-1)^k(c + 3c + c - 2).$$ Now choose $c$ so that $5c-2=0$ (so $c = \frac25$). That eliminates the awkward $(-1)^k$ term from the $b_k$ equation, which you should now be able to solve. Having found the answer for $b_k$, you then have $a_k = b_k - \frac25(-1)^k$.I need some help with this task. My theory book only shows examples of how to solve sequences in the form :

ππ = A * π(πβ1) β B * π(πβ2).

But I've no idea how to solve this task because of the alternating term. I've included the Answer (called "Svar") to the task.

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Thanks, that was a pretty solution! However the calculations got a bit messy while solving the characteristic equation for bk by hand so I went ahead and used a web calculator for it.

I'm going to give it another shot tomorrow solving it by hand.

We are not allowed to use a pocket calculator at the exam plus you lose a lot of credits if you go wrong somewhere in the calculations. That makes me a really angry student.

Edit: Solved the problem now! thanks again for the elegant solution.

I'm going to give it another shot tomorrow solving it by hand.

We are not allowed to use a pocket calculator at the exam plus you lose a lot of credits if you go wrong somewhere in the calculations. That makes me a really angry student.

Edit: Solved the problem now! thanks again for the elegant solution.

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