Question

What is the general solution of the recurrence relation \( a_n = 2a_{n-1} \) with \( a_0 = 5 \)?

• A. \( a_n = 5 \cdot 2^n \)
• B. \( a_n = 5 + 2^n \)
• C. \( a_n = 5n \)
• D. \( a_n = 2n \)

Hint:

Doubling each step means a geometric sequence. Write \( a_n = C \cdot 2^n \) and fix C using \( a_0 = 5 \).

Answer 1

The Correct Option is A

Step 1: This is a first order linear homogeneous recurrence. Each term is twice the previous one, so the sequence grows geometrically with ratio 2.

Step 2: Unroll it from the start:

\[ a_n = 2a_{n-1} = 2^2 a_{n-2} = \cdots = 2^n a_0 \]

Step 3: Substitute \( a_0 = 5 \).

\[ a_n = 5 \cdot 2^n \]

Check: \( a_0 = 5 \), \( a_1 = 10 \), \( a_2 = 20 \), matching the rule. The answer is option (A).