Problem Solving (PS) | Re: If a1 = 1 and a(n+1) – 3an + 2 = 4n for every positive integer n, then

Asked: If \(a_1 = 1\) and \(a_{n+1} – 3a_n+ 2 = 4n\) for every positive integer n, then \(a_{100}\) equals

In general
\(a_n = 3ˆn - 2n\)
\(a_{n+1} = 3ˆ(n+1) - 2(n+1)\)
\(a_{n+1} = 3(3ˆn - 2n) + (4n-2) = 3ˆ(n+1) - 6n + 4n - 2 = 3ˆ(n+1) - 2(n+1)\)

\(a_100 = 3ˆ100 - 2*100 = 3ˆ100 - 200\)

IMO C

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