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Now to find \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) First, it would be nice to know what \(a_0\) is, since plugging in \(n = 0\) simplifies the above formula greatly. In this case, \(a_0 = 2\) (work backwards from the sequence of constant differences). Thus

\begin{equation*} a_0 = 2 = a\cdot 0^2 + b \cdot 0 + c\text{,} \end{equation*}

so \(c = 2\text{.}\) Now plug in \(n =1\) and \(n = 2\text{.}\) We get

\begin{equation*} a_1 = 3 = a + b + 2 \end{equation*}

\begin{equation*} a_2 = 7 = a4 + b 2 + 2\text{.} \end{equation*}

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