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First, it is easy to check the initial condition: \(a_1\) should be \(2^1 + 1\) according to our closed formula. Indeed, \(2^1 + 1 = 3\text{,}\) which is what we want. To check that our proposed solution satisfies the recurrence relation, try plugging it in.

\begin{align*} 2a_{n-1} - 1 \amp = 2(2^{n-1} + 1) - 1\\ \amp = 2^n + 2 - 1\\ \amp = 2^n +1\\ \amp = a_n\text{.} \end{align*}

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