Question

1) a) Find the particular solution of the recurrence relation a(n+2) -4 a(n+1) + 4 (an) = 2n?
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Answer 1

Here is another take.

Let bn=2n. Then

an=2an−1+2bn−1,bn=2bn−1,b0=1

and so

(anbn)=(2022)(an−1bn−1)=2(1011)(an−1bn−1)

which gives

(anbn)=2n(1011)n(a0b0)=2n(10n1)(a0b0)

Therefore,

an=a02n+n2n