Answer: Let \(P_n\) be the statement \(u_n = 4\left(\frac{3}{4}\right)^n - 2\).
For \(n=1\),
\(4\left(\frac{3}{4}\right)^1 - 2 = 3 - 2 = 1\),
which agrees with \(u_1=1\). So \(P_1\) is true.
Now assume that \(P_k\) is true for some positive integer \(k\), so that
\(u_k = 4\left(\frac{3}{4}\right)^k - 2\).
Then, using the recurrence relation,
\(u_{k+1} = \frac{3u_k - 2}{4}\)
\(= \frac{3\left(4\left(\frac{3}{4}\right)^k - 2\right) - 2}{4}\)
\(= \frac{12\left(\frac{3}{4}\right)^k - 8}{4}\)
\(= 3\left(\frac{3}{4}\right)^k - 2\)
\(= 4\left(\frac{3}{4}\right)^{k+1} - 2\),
since \(4\left(\frac{3}{4}\right)^{k+1} = 3\left(\frac{3}{4}\right)^k\).
Therefore \(P_{k+1}\) is true whenever \(P_k\) is true.
Hence, by mathematical induction, \(u_n = 4\left(\frac{3}{4}\right)^n - 2\) for all positive integers \(n\).
Let \(P_n\) denote the proposition \(u_n = 4\left(\frac{3}{4}\right)^n - 2\).
Base case: When \(n=1\),
\(4\left(\frac{3}{4}\right)^1 - 2 = 3 - 2 = 1\).
Since \(u_1=1\), \(P_1\) is true.
Inductive step: Assume \(P_k\) is true for some positive integer \(k\). Then
\(u_k = 4\left(\frac{3}{4}\right)^k - 2\).
Using the recurrence relation,
\(u_{k+1} = \frac{3u_k - 2}{4}\).
Substitute the inductive hypothesis:
\(u_{k+1} = \frac{3\left(4\left(\frac{3}{4}\right)^k - 2\right) - 2}{4}\)
\(= \frac{12\left(\frac{3}{4}\right)^k - 6 - 2}{4}\)
\(= \frac{12\left(\frac{3}{4}\right)^k - 8}{4}\)
\(= 3\left(\frac{3}{4}\right)^k - 2\).
Now
\(3\left(\frac{3}{4}\right)^k = 4\cdot \frac{3}{4}\left(\frac{3}{4}\right)^k = 4\left(\frac{3}{4}\right)^{k+1}\),
so
\(u_{k+1} = 4\left(\frac{3}{4}\right)^{k+1} - 2\).
This is exactly \(P_{k+1}\).
Therefore, if \(P_k\) is true then \(P_{k+1}\) is true.
Since \(P_1\) is true and \(P_k \Rightarrow P_{k+1}\), it follows by induction that \(u_n = 4\left(\frac{3}{4}\right)^n - 2\) for all positive integers \(n\).
