# Is my short induction proof correct?

## Homework Statement

"Prove: ##∀n∈ℕ##, ##3^n>n^2##

## The Attempt at a Solution

(1) We will prove that ##3^n>n^2## at ##n=1##
##3=3^1>1=1^2##

(2) Now assume that ##3^k>k^2## for some ##k>1##
(3) We will prove that ##3^{k+1}>(k+1)^2## or ##3⋅3^k>k^2+2k+1##
Note that ##k^2+2k^2+1=3k^2+1≥k^2+2k+1##, and as such, ##3⋅3^k≥3k^2+3>3k^2+1≥k^2+2k+1##. So ##3^{k+1}>3k^2+1≥(k+1)^2##.

I'm basically ambivalent as to whether to insert that extra term in the inequality: ##3k^2+1##, not knowing whether it is equal to, less than, or greater than ##3^{k+1}##. Can someone check my work, please? Thanks.

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Mark44
Mentor

## Homework Statement

"Prove: ##∀n∈ℕ##, ##3^n>n^2##

## The Attempt at a Solution

(1) We will prove that ##3^n>n^2## at ##n=1##
##3=3^1>1=1^2##

(2) Now assume that ##3^k>k^2## for some ##k>1##
(3) We will prove that ##3^{k+1}>(k+1)^2## or ##3⋅3^k>k^2+2k+1##
Note that ##k^2+2k^2+1=3k^2+1≥k^2+2k+1##, and as such, ##3⋅3^k≥3k^2+3>3k^2+1≥k^2+2k+1##. So ##3^{k+1}>3k^2+1≥(k+1)^2##.

I'm basically ambivalent as to whether to insert that extra term in the inequality: ##3k^2+1##, not knowing whether it is equal to, less than, or greater than ##3^{k+1}##. Can someone check my work, please? Thanks.
The induction step boils down to showing that ##3k^2 > (k + 1)^2##. It's easy enough to prove the equivalent statement ##3k^2 - (k + 1)^2 > 0##.

Let's see...

##3k^2-k^2-2k-1=2k^2-2k-1=2k(k-1)-1>0## for ##k>0##?

Mark44
Mentor
Let's see...

##3k^2-k^2-2k-1=2k^2-2k-1=2k(k-1)-1>0## for ##k>0##?
Almost. 2k(k - 1) - 1 > 0 for k > 1.
If k = 0, you have 0(-1) - 1 < 0, and if k = 1, you have 2(0) - 1 < 0.