We have
$P(n): b_n=5+4 n$, for all natural numbers given that $b_0=$ 5 and $b_k=4+b_{k-1}$
For $\mathrm{n}=1$
$P(1): b_1=5+4 \times 1=9$
As $\quad b_0=5, b_1=4+b_0=4+5=9$
Hence, $P(1)$ is true.
Now, let $P(n)$ is true for $n=k$.
$$
P(k): b_k=5+4 k
$$
Now, to prove $P(k+1)$ is true, we have to show that
$$
\begin{aligned}
\therefore \quad P(k+1): b_{k+1} &=5+4(k+1) \\
\text { Now, } \quad b_{k+1} &=4+b_{k+1-1} \text { (given) } \\
&=4+b_k=4+5+4 k=5+4(k+1)
\end{aligned}
$$
So, by the mathematical induction $P(k+1)$ is true whenever $P(k)$ is true, hence $P(n)$ is true.