Can you use mathematical induction to prove that #2^n > 4n# for all #n in ZZ^+ n>=5#?

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Answer 1 · 2017-02-03T06:38:57

Proof: (by induction)

Base case: For #n=5#, we have #2^5 = 32 > 20 = 4(5)#.

Inductive hypothesis: Suppose that #2^k > 4k# for some integer #k>=5#.

Induction step: We wish to show that #2^(k+1) > 4(k+1)#. Indeed,

#2^(k+1) = 2*2^k#

#=2^k+2^k#

#>4k+4k" "# (by the inductive hypothesis)

#>4k+4" "# (as #k>=5#)

#=4(k+1)#

We have supposed true for #k# and shown true for #k+1#. Thus, by induction, #2^n>4n# for all integers #n>=5#.
∎