<s>[INST] Prove using mathematical induction Sequence $b_1, b_2,...$ where $b_1=b_2=1$ and for every natural number $k$ bigger than $2$, $b_k = b_{k-1}+b_{k-2}$.

Prove using mathematical induction that, for every $k \in \mathbb N, b_k \le (\frac{7}{4})^{k-1} $


I started with if  $k=3$, $k_3=1+1=2$ and $(\frac{7}{4})^{k-1} =(\frac{7}{4})^2=3 \frac{1}{16}$, $2<3 \frac{1}{16}$. This is the induction base?


Step:
Let us assume that S(m) is true, $b_m \le (\frac{7}{4})^{m-1}$ for every m $\in$ {3, ..., k}.

We have to show that  $b_{k+1} \le (\frac{7}{4})^k$.
Using assumption and $b_{k+1} = b_{k}+b_{k-1}$ I got
$$b_k+b_{k-1}\le (\frac{7}{4})^{k+1}+(\frac{7}{4})^{k}$$
I don't know what to do next.
 [/INST] 
I don't know what to do next.

$ \left(\dfrac{7}{4}\right)^{k+1}+\left(\dfrac{7}{4}\right)^{k}=\left(\dfrac74\right)^k\left(\dfrac74+1\right)=\left(\dfrac74\right)^k\left(\dfrac{11}4\right)=\left(\dfrac74\right)^k\left(\dfrac{44}{16}\right)$
$<\left(\dfrac74\right)^k\left(\dfrac{49}{16}\right)=\left(\dfrac74\right)^k\left(\dfrac{7}{4}\right)^2=\left(\dfrac74\right)^{k+2}$
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