1)Given

\(u_n+1=7-2u_n\)

\(u_2=5\)

Determine \(u_1\)

Let us evaluate the recurrence relation \(u_n+1=7-2u_n\) at n=1 \(u_2=7-2u_1\)

Since \(u_2=5\) has been given: \(5=7-2u_1\)

Subtract 7 from each side: \(5-7=7-2u_1-7\)

Combine like terms: \(-2=-2u_1\)

Divide each side by -2: \(\frac{-2}{-2}=\frac{-2u_1}{-2}\)

Evaluate: \(1=u_1\)

Thus we then obtained \(u_1=1\)

2)Determine \(u_0\)

Let us evaluate the recurrence relation \(u_n+1=7-2u_n\) at n=0 u\(1=7-2u_0\)

Since \(u_2=5\) has been given: \(1=7-2u_0\)

Subtract 7 from each side: \(1-7=7-2u_0-7\)

Combine like terms: \(-6=-2u_0\)

Divide each side by -2: \(\frac{-6}{-2}=\frac{-2u_0}{-2}\)

Evaluate: \(3=u_0\)

Thus we then obtained \(u_0=3\)