Prove that for each natural number prove n≥2, 2^n+1 ≤ 3^n

Question

foundation of mathematics 

Answer 1

I am guessing that you want to prove that for n≥2, 2⁽ⁿ⁺¹⁾ ≤ 3ⁿ because the other proposition seems less challenging.

It is validated when n=2 because 2³ < 3².

 

For other cases, n>2, we examine 2ⁿ⁺¹ : 3² using the colon (:) as an unknown inequality symbol.

Equivalently, we are investigating  2•2ⁿ : 3ⁿ which can be rewritten as 2 : (3/2)ⁿ for n > 2.

 

2 < (1.5)ⁿ when n= 2 or more because the right side is at least 2.25

 

That proves it for me!