Solution to Solve the recurrence relation an=6 an-2-12 an-2 + 8 an-2 + n2 2n - Sikademy
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Archangel Macsika

Solve the recurrence relation an=6 an-2-12 an-2 + 8 an-2 + n2 2n

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Let us solve the recurrence relation a_n=6 a_{n-2}-12a_{n-2} + 8a_{n-2} + n^2+ 2n, which is equivalent to a_n-2a_{n-2}= n^2+ 2n. The characteristic equation k^2-2=0 of the homogeneous recurence relation a_n-2a_{n-2}=0 has the roots k_1=\sqrt{2},\ k_2=-\sqrt{2}.

It follows that the solution of a_n-2a_{n-2}= n^2+ 2n is of the form a_n=C_1(\sqrt{2})^n+C_2(-\sqrt{2})^n+b_p(n), where b_p(n)=an^2+bn+c. Then an^2+bn+c-2(a(n-2)^2+b(n-2)+c)=n^2+2n, and hence an^2+bn+c-2(an^2-4an+4a+bn-2b+c)=n^2+2n. It follows that -an^2+(8a-b)n-8a+4b-c=n^2+2n, and thus a=-1,\ 8a-b=2,\ -8a+4b-c=0. Therefore, a=-1,\ b=-10,\ c= -32.


We conclude that the solution is the following:

a_n=C_1(\sqrt{2})^n+C_2(-\sqrt{2})^n-n^2-10n-32.

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