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: For a recurrence of the form $a_n = r\,a_{n-1}$, the solution is $a_n = C\,r^{\,n}$, where $C$ is fixed by the initial value. Here $r = 2$.

Step 2: Use the initial condition at $n = 0$. Since $r^0 = 1$, we get $a_0 = C = 5$.

Step 3: Plug the constant back in.

\[ a_n = 5 \cdot 2^{\,n} \]

\[ \boxed{a_n = 5 \cdot 2^n} \]