Consider the nonhomogeneous linear recurrence relation an = 3an−1 + 2^n. Show
that a^n = – 2^(n+1) is a solution of this recurrence relation.
an−1=−2na_{n-1}=-2^nan−1=−2n
then:
3an−1+2n=−3⋅2n+2n=2n(1−3)=−2⋅2n=−2n+1=an3a_{n−1} + 2^n=-3\cdot2^n+2^n=2^n(1-3)=-2\cdot2^n=-2^{n+1}=a_n3an−1+2n=−3⋅2n+2n=2n(1−3)=−2⋅2n=−2n+1=an
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