Question

Prove that $3^{n+1}>3(n+1)$.

Answer 1

Let P(n): $3^{n+1}$ > 3(n + 1)
Step 1: Put n = 1
Then, 3${}^2$ > 3(2)
$\Rightarrow$ p(n) is true for n = 1
Step 2: Assume that p(n) is true for n = k then, $3^{k+1}$ > 3(k + 1)
Multiplying throughout with '3'.
$3^{k+1}\times 3>3(k+1)\times 3=9k+9=3(k+2)+(6k+3)>3(k+2)$
$\Rightarrow 3^{\overline{k+1}+1}>3(\overline{k+1}+1)$
P(n) is true for n = k + 1
$\therefore$ By the principle of mathematical induction,
P(n) is true for all n $ϵ$ N.
Hence, $3^{n+1}$ > 3(n + 1) $\forall$ n $ϵ$ N.