Question:
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ is defined by letting $a_{1}=3$ and $a_{k}=7 a_{k-1}$ for all natural numbers $k \geq 2$. Show that $a_{n}=3 \cdot 7^{n-1}$ for all $n \in \mathbf{N}$.
Solution:
Let $\mathrm{P}(n): a_{n}=3 \cdot 7^{n-1}$ for all $n \in \mathbf{N}$.
Step I: For $n=1$,
$P(1)$ :
$a_{1}=3 \cdot 7^{1-1}=3 \cdot 1=3$
So, it is true for $n=1$.
Step II : For $n=k$,
Let $\mathrm{P}(k): a_{k}=3 \cdot 7^{k-1}$ be true for some $k \in \mathbf{N}$ and $k \geq 2$.
Step III : For $n=k+1$,
$a_{k+1}=7 a_{k}$
$=7 \cdot 3 \cdot 7^{k-1}$ (Using step II)
$=3 \cdot 7^{k-1+1}$
$=3 \cdot 7^{(k+1)-1}$
So, it is also true for $n=k+1$.
Hence, $a_{n}=3 \cdot 7^{n-1}$ for all $n \in \mathbf{N}$.