Question

If \( a_1 = 1 \) and \( a_{n+1} = 2a_n + 5 \), n = 1, 2, ..., then \( a_{100} \) is equal to:

• A. \( (6 \ast 2^{99} - 5) \)
• B. \( (5 \ast 2^{99} + 6) \)
• C. \( (6 \ast 2^{99} + 5) \)
• D. \( (5 \ast 2^{99} - 6) \)

Hint:

Add 5 to both sides of the recurrence so that it becomes a pure doubling relation.

Answer 1

The Correct Option is A

Step 1: Rewrite the recurrence:
\( a_{n+1} = 2a_n + 5 \) gives \( a_{n+1} + 5 = 2a_n + 10 = 2(a_n + 5) \).

Step 2: Recognise a geometric progression:
Let \( b_n = a_n + 5 \). Then \( b_{n+1} = 2b_n \), with \( b_1 = a_1 + 5 = 6 \). So \( b_n = 6 \cdot 2^{n-1} \).

Step 3: Go back to \( a_n \):
\( a_n = 6 \cdot 2^{n-1} - 5 \). For \( n = 100 \): \( a_{100} = 6 \cdot 2^{99} - 5 \).

Step 4: Check with small n:
\( a_2 = 2(1) + 5 = 7 \) and the formula gives \( 6 \cdot 2 - 5 = 7 \). \( a_3 = 19 \) and the formula gives \( 6 \cdot 4 - 5 = 19 \).

Final Answer:
The correct option is (1): \( 6 \ast 2^{99} - 5 \)