Solution 2.4.3.1.

\(a_n = 3 + 2^{n+1}\text{.}\) We should use telescoping or iteration here. For example, telescoping gives

\begin{equation*} \begin{aligned} a_1 - a_0 \amp = 2^1\\ a_2 - a_1 \amp = 2^2\\ a_3 - a_2 \amp = 2^3\\ \vdots\amp \vdots\\ a_n - a_{n-1} \amp = 2^n \end{aligned} \end{equation*}

which sums to \(a_n - a_0 = 2^{n+1} - 2\) (using the multiply-shift-subtract technique from Section 2.2 for the right-hand side). Substituting \(a_0 = 5\) and solving for \(a_n\) completes the solution.

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