Question
A sequence $a_1, a_2, a_3,$ ... is defined by letting $a_1 = 3$ and $a_k = 7a_k - 1$ for all natural numbers $\text{k}\geq2.$ Show that $a_n = 3.7^{n-1}$ for all $\text{n}\in\text{N}.$

Answer

Let $p(n)$ be the statement given by $P(n): a_n=3 \times 7^{n-1}$ for all $\mathbf{n} \in \mathbf{N}$.
Step 1: $p(2): a_2=3 \times 7^{2-1}=21$ Given that $a_k=7 a_{k-1}$ for all natural numbers $k \geq 2 a_2=7 a_1=7 \times 3=21$
$\therefore p(2)$ is true. Step 2: Let $p(m)$ is true.
Then, $a_m=3 \times 7^{m-1}$ ... (1)
We have to prove that $p(m+1)$ is true.
$a_{m+1}=7 a_m a_{m+1}=7 \times a_m a_{m+1}=7^1 \times 3 \times 7^{m-1} \ldots$ [From (1)] $a_{m+1}=3 \times 7^{m-1+1} a_{m+1}=3 \times 7^m \Rightarrow p(m+1)$ is true.
Hence by the principle of mathematical induction, the given results is true for all $\mathbf{n} \in \mathbf{N}$.

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