Induction (really simple) (1 Viewer)

Timothy.Siu

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Prove by induction:
2^n > 3n^2 for n => 8
or 2^n-3n^2>0
n=8 LHS>0

assume n=k is true, 2^k-3k^2>0

prove n=k+1
i.e. 2^(k+1) -3(k+1)^2 >0

LHS=2.2^k-3(k+1)^2
=2(2^k-3k^2)+6k^2-3(k+1)^2
> 6k^2-(3k^2+6k+3)=3k^2-6k-3=3(k^2-2k-1)=3(k-1)^2-6
which is >0 for n>=8
 

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