prove using induction that for all natural number $n≥2$, $3^n>3n+1$

Question

(Answered)

1. base $$3^2>3(2)+1$$ $$9>6+1$$ $$9>7$$
   which is true

2. assume $k=n$ $3^k>3n+1$
   should be true for n=k+1
   $$3^{k+1}>3(k+1)+1=3k+4$$ $$3^{k+1} $$ $$=3^k*3$$ $$\gt (3k+1)*3$$ $$=9k+3$$ $$\gt3k+4$$ because k is a natural number

Is my proof correct?

Answer 1

You want to prove that $3^{k+1}>3(k+1)+1$, so it's not very good to write this inequality from the beginning. You want to end with it. Start from the left hand side only:

$3^{k+1}=3\cdot 3^k>3(3k+1)=9k+3=3k+(6k+3)\geq3k+4=3(k+1)+1,$

since clearly $6k+3\geq 4$.