Question

A sequence $a_1,a_2,a_3,\cdots$ is defined by letting $a_1=3$ and $a_k=7a_{k-1}$ for all natural numbers $k\ge 2.$ Then $a_n=$
(Solve using mathematical induction)

• A. $7^n$
• B. $3\cdot 7^n$
• C. $7^{n-1}$
• D. $3\cdot 7^{n-1}$

Answer 1

The correct option is D $3\cdot 7^{n-1}$

Given : $a_1=3$ and $a_k=7a_{k-1}$
From mathematical induction property : sequence is true for all $k\in N$
For $k=2,a_2=7a_1=3\cdot 7$
For $k=3,a_3=7a_2=7(3\cdot 7)=3\cdot 7^2$
Similarly, let for $k=n,a_n=3\cdot 7^{n-1}$ is true
now check at $k=n+1$
$\Rightarrow a_{n+1}=7a_n=7(3\cdot 7^{n-1})$
$\Rightarrow a_{n+1}=3\cdot 7^n=3\cdot 7^{(n+1)-1}$
As, $a_n$ is also true for $k=n+1$
$\therefore a_n=3\cdot 7^{n-1},n\ge 2$