Recurrence induction show $f(n) = c4^{n} - d2^{n}, \forall n \in \mathbb N$

Question

Use induction to find positive $c, d \in \mathbb R$ such that $f(n) = c4^{n} - d2^{n}, \forall n \in \mathbb N$

My attempt!

Base Case:

Let $(n = 0)$

$f(n) = 5$ [Def of f] and $c4^n - d2^n = c - d$

Therefore we need $5 = c - d$ (*)

Inductive step: Suppose $f(n) = c4^n - d2^n$ [I.H]

Trying to show: $f(n+1) = c4^{n+1} - d2^{n+1}$

$f(n+1) = 4f(n) + 3 \cdot 2^{n}$ [def of f; $n>0$]

$= c4^{n+1} - 4d2^n + 3 \cdot 2^n$ [I.H]

$= c4^{n+1} - 2^{n+1} (2d - \frac{3}{2})$ [Algebra]

Confused at this part. How do I find the c, d, and satisfy $(*)$?

Answer 1

Here is a roadmap:

1. Find $c,d$ by computing $f(0)$ and $f(1)$.

2. Prove by induction that $f(n) = c4^{n} - d2^{n}$ satisfies the recurrence.