Prove or disprove: 5^n + 9 < 6^n for all n >= 2

flakine

Junior Member
Joined
Aug 24, 2005
Messages
78
\(\displaystyle & \text{Prove or disprove the statement: }\)
\(\displaystyle & \text{ 5}^n + 9 < 6^n ,\text{ }\forall \text{ integers n; n} \geqslant \text{2}\)

\(\displaystyle & \text{Show p(a) ['a' is min value of n]}\)
\(\displaystyle & \text{ 5}^2 + 9 < 6^2\)
\(\displaystyle & \text{ 25} + 9 < 36\)
\(\displaystyle & \text{ 34 < 36 [so p(a) is true]}\)

\(\displaystyle & \text{Suppose p(k); k = n}\)
\(\displaystyle & \text{ 5}^k + 9 < 6^k \text{ [inductive step]}\)

\(\displaystyle & \text{We wish to show p(k + 1)}\)
\(\displaystyle & \text{ 5}^{k + 1} + 9 < 6^{k + 1}\)

\(\displaystyle & \text{ Starting on left - hand side}\)
\(\displaystyle & \text{ 5}^{k + 1} + 9 = 5(5^k ) + 9\)
\(\displaystyle & \text{ = ??????} \cr}\)

Having a tough time with closing these types of problem. Any help would be appreciated.
 
Re: Proving an Inequality

\(\displaystyle 6^{k+1} = 6(6^k) >_{(i.h.)} 6(5^k+9) > 5(5^k+9) = 5^{k+1}+45 > 5^{k+1}+9\)
 
Re: Proving an Inequality

How did \(\displaystyle 5(5^k ) + 9\) become \(\displaystyle 5(5^k + 9)\) ?
 
Re: Proving an Inequality

flakine said:
How did \(\displaystyle 5(5^k ) + 9\) become \(\displaystyle 5(5^k + 9)\) ?

...it didn't?
 
Top