Question

Let b0, b1, b2, be the sequence defined by the explicit formula bn = c · 3n d(-2)n for every integer n ≥ 0, where c and d are real numbers. Fill in the blanks in the following proof to show that for any choice of c and d and for each integer k ≥ 2, bk = bk - 1 6bk - 2. Proof: Let c and d be any real numbers and let k be any integer such that k ≥ 2. By definition of b0, b1, b2, ________.

Answer 1

To prove the recursive sequence relation for bk = bk - 1 + 6bk - 2, we follow inductive reasoning by writing out the explicit terms and then demonstrating the relation for k+1 based on the assumption it holds for k, showing the relation holds for all k >= 2.

Step-by-step explanation:

To show that the sequence defined by bn = c \cdot 3n + d(-2)n for every integer n \geq 0, satisfies the relation bk = bk-1 + 6bk-2 for all integers k \geq 2, we start by writing down the terms explicitly.

By definition of b0, b1, b2:

• b0 = c \cdot 30 + d(-2)0 = c + d
• b1 = c \cdot 31 + d(-2)1 = 3c - 2d
• b2 = c \cdot 32 + d(-2)2 = 9c + 4d

We assume the relation holds for some integer k, and then demonstrate it for k+1. This utilizes inductive reasoning to show the recursive nature of the sequence:

1. bk = c \cdot 3k + d(-2)k
2. bk-1 = c \cdot 3k-1 + d(-2)k-1
3. 6bk-2 = 6[c \cdot 3k-2 + d(-2)k-2]

To prove bk+1 using the inductive step, we check:

• bk+1 = c \cdot 3k+1 + d(-2)k+1
• bk+1 = 3bk - 2dbk-1
• bk+1 = 3[c \cdot 3k + d(-2)k] - 2d[c \cdot 3k-1 + d(-2)k-1]
• bk+1 = bk + 6bk-2

Thus, the relation holds for all k \geq 2.