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This closed formula would have \(a_{n-1} = 3\cdot 2^{n-1} + 7 \cdot 5^{n-1}\) and \(a_{n-2} = 3\cdot 2^{n-2} + 7 \cdot 5^{n-2}\text{.}\) Then we would have
\begin{align*}
7a_{n-1} - 10a_{n-2} = \amp 7(3\cdot 2^{n-1} + 7 \cdot 5^{n-1}) - 10(3\cdot 2^{n-2} + 7 \cdot 5^{n-2})\\
= \amp 21\cdot 2^{n-1} + 49 \cdot 5^{n-1} - 30\cdot 2^{n-2} - 70 \cdot 5^{n-2})\\
= \amp 21\cdot 2^{n-1} + 49 \cdot 5^{n-1} - 15\cdot 2^{n-1} - 14 \cdot 5^{n-1})\\
= \amp 6\cdot 2^{n-1} + 35 \cdot 5^{n-1}\\
= \amp 3\cdot 2^{n} + 7 \cdot 5^{n} = a_n\text{.}
\end{align*}
So the closed formula agrees with the recurrence relation. The closed formula has initial terms \(a_0 = 10\) and \(a_1 = 41\text{.}\)
in-context