Step-by-step explanation:
(a) Use the following recursive definition to get the next four terms in the sequence a3, a4, a5, a6:
a3 = 2a1 + a2 = 2(1) + 3 = 2 + 3 = 5
a4 = 2a2 + a3 = 2(3) + 5 = 6 + 5 = 11
a5 = 2a3 + a4 = 2(5) + 11 = 10 + 11 = 21 a6 = 2a4 + a5 = 2(11) + 21 = 22 + 21 = 43
As a result, the sequence's next four terms are a3 = 5, a4 = 11, a5 = 21, and a6 = 43.
(b) One property that may apply to all k 1 is that the sequence ak is an integer.
(c) To demonstrate the property that the sequence ak is an integer for all k 1, we'll need to present two base cases and the inductive step:
**Standard Cases:**
1. When k = 1, a1 equals 1, which is an integer.
2. a2 = 3 for k = 2, which is also an integer.
**Inductive Step:** Assume that aj is an integer for any integers j such that 1 j n. We'd like to demonstrate that an+1 is also an integer.
Using the sequence's recursive definition: an+1 = 2an-1 + an
By the inductive hypothesis, an-1 and an are both integers because we believed that aj is an integer for 1 j n. As a result, 2an-1 is an integer (since the product of two integers is an integer). Adding an integer (an) to another integer (2an-1) yields another integer, demonstrating that an+1 is an integer.
We demonstrated that the property "ak is an integer for all k 1" is true for this sequence using the strong induction method.