Prove that 2n>4n for n≥5.
Prove for n=5n=5n=5 :
25=32>20=4⋅52^5=32>20=4\cdot525=32>20=4⋅5
We prove by induction, let be true for n−1n-1n−1
2n=2⋅2n−1>True for n-12⋅4(n−1)=4n+(4n−8)2^n=2\cdot2^{n-1}\stackrel{\text{True for n-1}}{>}2\cdot4(n-1)=4n+(4n-8)2n=2⋅2n−1>True for n-12⋅4(n−1)=4n+(4n−8)
Because 4n−8>04n-8>04n−8>0 for n≥5n\geq5n≥5 :
4n+(4n−8)>4n4n+(4n-8)>4n4n+(4n−8)>4n
So, we prove that 2n>4n2^n>4n2n>4n for n≥5n\geq5n≥5
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