Trouble solving this induction problem

Question

Show that, for every $n\ge2$, $3^n >n(n-1)$.

Well, I started by showing the base case ($n = 2$): $3^2 > 2$

Now, for $n+1$: $P(n)\Rightarrow P(n+1)$ $$3^{n+1} > (n+1)n$$

My solution is this: $$3^{n+1} = 3^{n}\times 3 > 3^n \times 1 > 3^n \times n$$(because $n\ge 2$) > $(n+1)n$ (by the hypothesis)

Is this reasoning correct? What other paths could I've taken to solve this problem?

Answer 1

Show that, for every $n\geq2, 3^n >n(n-1)$.

Well, I started by showing the base case($n = 2$): $3^2 > 2$

Now assume that the statement we are attempting to prove is actually true up to $n$; we claim that $P(n+1) = 3^{n+1} > (n+1)(n)$ is true.

Well, on the LHS you get $(3^n)(3) > (n)(n-1)(3)$ (true from our assumption) but for all $n > 2$ the RHS is greater than $(n)(n+1)$. So our final statement looks like $$(3^n)(3) > (n)(n-1)(3) > (n+1)(n)$$

Answer 2

we have $$3^n>n(n-1)$$ and we have to prove that $$3^{n+1}>n(n+1)$$ for $n\geq 2$ is hold.Multilying the first inequality by $3$ we get $$3^{n+1}>3n(n-1)$$ we have to show that $$3n(n-1)\geq n(n+1)$$ which is equivalent to $$n\geq 2$$