Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n < \left(\frac{7}{4}\right)^n$ for all natural numbers.

Question

I assume I'm supposed to use induction. base step is easy. I'm stuck on how to form the inductive step. Any tips are greatly appreciated.

After showing that it's true for $n=1,2$ then note that: $a_{n}=a_{n-1}+a_{n-2}<(\frac{7}{4})^{n-1}+(\frac{7}{4})^{n-2}=(\frac{7}{4})^{n-2}(\frac{11}{4})=(\frac{7}{4})^{n}(\frac{44}{49})<(\frac{7}{4})^{n}$. — user71352, Jun 15 '14 at 21:21
I could but I'm going to let someone else take it. I'm happy to just help the OP. — user71352, Jun 15 '14 at 21:26
@clay Since no one posted an answer, I suggest you do it, so this question doesn't come up as unanswered. — Git Gud, Jul 11 '14 at 13:58

score 1 · Accepted Answer · answered Jul 11 '14 at 20:57

Here is the inductive step for anyone reading this...

Assuming $P(k-2)$: $a_{k-2} < \left(\frac{7}{4}\right)^{k-2}$

Assuming $P(k-1)$: $a_{k-1} < \left(\frac{7}{4}\right)^{k-1}$

Definition of $a_k$: $a_k=a_{k-1}+a_{k-2}$

Combining with inductive assumptions: $a_k < \left(\frac{7}{4}\right)^{k-2} + \left(\frac{7}{4}\right)^{k-1}$

Algebraically factor out $\left(\frac{7}{4}\right)^{k-2}$: $a_k < \left(\frac{7}{4}\right)^{k-2} \cdot \left(1 + \frac{7}{4}\right)$

$a_k < \left(\frac{7}{4}\right)^{k-2} \cdot \frac{11}{4} = \left(\frac{7}{4}\right)^k \cdot \left(\frac{4}{7}\right)^2 \cdot \frac{11}{4} = \left(\frac{7}{4}\right)^k \cdot \frac{44}{49} < \left(\frac{7}{4}\right)^k$

$a_k < \left(\frac{7}{4}\right)^k$

Let $a_1=1$, $a_2=3$ , and for $n \ge 2$ let $a_n=a_{n-1}+a_{n-2}$. Show that $a_n < \left(\frac{7}{4}\right)^n$ for all natural numbers.

1 Answers1

Linked