Proof by induction: 5^n + 9 < 6^n for all integers n≥2

[savtaylor2010]

Homework Statement

Prove the statement by mathematical induction:
5ⁿ + 9 < 6ⁿ for all integers n≥2

Homework Equations

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The Attempt at a Solution

Proof: let P(n) be the statement,
5ⁿ + 9 < 6ⁿ

P(2) is true because,
34<36.

Suppose that P(n) is true.
P(n+1) would be,
5ⁿ⁺¹ + 9 < 6ⁿ⁺¹
6ⁿ⁺¹ = 6$\bullet$6ⁿ
6$\bullet$(5ⁿ+9)< 6$\bullet$6ⁿ
==> 54+ 6(5ⁿ) < 6$\bullet$6ⁿ
==> 54+ 5(5ⁿ) + 5ⁿ < 6$\bullet$6ⁿ
==> 54+ 5ⁿ⁺¹ + 5ⁿ < 6$\bullet$6ⁿ

What do I do from here? I don't know how to wrap up and prove this induction, or that P(n+1) is real? Any help would be much appreciated!

 

[micromass]

It seems you need to show that $9<54+5^n$. That can't be too difficult.

 

[savtaylor2010]

I don't really understand the hint that you gave. Could you elaborate a little more on that? I just have a hard time understanding the structure of inequality inductions.

 

[Office_Shredder]

Your confusion seems to stem from the fact that you write down what you're trying to prove, and then write down something else entirely and lose track of what your final line needs to look like

First,
6(5ⁿ+9) < 6*6ⁿ

This is true by the inductive hypothesis assuming that P(n) is true. From here you conclude that
54+ 5ⁿ⁺¹ + 5ⁿ < 6ⁿ⁺¹

Our objective is to get
5ⁿ⁺¹+9<6ⁿ⁺¹

So it suffices to show that
5ⁿ⁺¹+9 < 54+5ⁿ⁺¹+5ⁿ

 

[savtaylor2010]

Okay. That makes sense now. I guess it takes some practice for it to become more clear. I see now that you manipulated one side of the inequality, then related it back to it's original p(n+1) state to prove that it is in fact less than the other side of the inequality.