A abdullion New Member Joined Mar 18, 2015 Messages 12 Gender Male HSC 2016 Oct 23, 2015 #1 Please help, c: - I don't know how to do it, so if someone could show the step by step working it'd be great. 10^n>=7^n+5^n for all integers n>=2
Please help, c: - I don't know how to do it, so if someone could show the step by step working it'd be great. 10^n>=7^n+5^n for all integers n>=2
G glittergal96 Active Member Joined Jul 25, 2014 Messages 418 Gender Female HSC 2014 Oct 23, 2015 #2 Let P(n)= 10^n - 7^n - 5^n. Then P(2)=100 - 49 - 25 > 0. Now suppose P(n) > 0. Then P(n+1)=10^(n+1)-7^(n+1)-5^(n+1) >10^(n+1)-10*7^n-10*5^n =10*P(n) >0. Hence P(n) > 0 for all positive integers n >= 2 by induction.
Let P(n)= 10^n - 7^n - 5^n. Then P(2)=100 - 49 - 25 > 0. Now suppose P(n) > 0. Then P(n+1)=10^(n+1)-7^(n+1)-5^(n+1) >10^(n+1)-10*7^n-10*5^n =10*P(n) >0. Hence P(n) > 0 for all positive integers n >= 2 by induction.
