Solve the following
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 $n \in \mathbf{N}$.