Chapter 5 - Section 5.3 - Recursive Definitions and Structural Induction - Exercises - Page 358: 7

Part A: $a_1 = 6$, $a_{n+1} = a_n + 6$. Part B: $a_1 = 3$, $a_{n+1} = a_n + 2$. Part C: $a_1 = 10$, $a_{n+1} = 10a_n$. Part D: $a_1 = 5$, $a_{n+1} = a_n$.

Part A: The sequence generated by this recursion begins with $6,12,18,\ldots$ which is simply the multiples of 6. Since you can obtain the multiples of 6 by successively adding 6, we can write $a_1 = 6$, $a_{n+1} = a_n + 6$ as a recursive relation for this sequence. Part B: The sequence generated by this recursion begins with $3,5,7,\ldots$ which is the sequence of odd integers starting from 3. Since each odd integer differs from another by 2, we can write $a_1 = 3$, $a_{n+1} = a_n + 2$ as a recursive relation for this sequence. Part C: The sequence generated by this recursion begins with $10,100,1000,\ldots$ which is the sequence of powers of 10. Since each power of ten differs from another by a factor of 10, we can write $a_1 = 10$, $a_{n+1} = 10a_n$ as a recursive relation for this sequence. Part D: The sequence generated by this recursion begins with $5,5,5,\ldots$ which is a sequence where every term equals 5. Since all terms are the same, we can write $a_1 = 5$, $a_{n+1} = a_n$ as a recursive relation for this sequence.