b) \( t(1)=3, t(n+1)=t(n)+6 \)

To solve the recurrence relation given by \( t(1) = 3 \) and \( t(n+1) = t(n) + 6 \), we will find a general formula for \( t(n) \). ### Step 1: Identify the pattern We can calculate the first few terms to identify a pattern: - For \( n = 1 \): \[ t(1) = 3 \] - For \( n = 2 \): \[ t(2) = t(1) + 6 = 3 + 6 = 9 \] - For \( n = 3 \): \[ t(3) = t(2) + 6 = 9 + 6 = 15 \] - For \( n = 4 \): \[ t(4) = t(3) + 6 = 15 + 6 = 21 \] From these calculations, we can see that the sequence is increasing by 6 each time. ### Step 2: General formula The pattern suggests that \( t(n) \) can be expressed as: \[ t(n) = t(1) + (n-1) \cdot 6 \] Substituting \( t(1) = 3 \): \[ t(n) = 3 + (n-1) \cdot 6 \] ### Step 3: Simplify the formula Now, we can simplify the formula: \[ t(n) = 3 + 6n - 6 = 6n - 3 \] ### Conclusion The general formula for the sequence defined by the recurrence relation is: \[ t(n) = 6n - 3 \] This formula allows us to calculate \( t(n) \) for any positive integer \( n \).