Question #189930

We do a proof by induction to show the following inequality is true for all n∈N: 4n ≥ 4n

Let P(n) represent the statement: 4n ≥ 4n

For n=1: LHS = 41 = 4 and RHS = 4(1) = 4

Since 4≥4, P(1) is true.

Assume true for n=k: P(k)true means 4k ≥ 4k

Show true for n=k+1 assuming P(k):

Provide the correct proof.








1
Expert's answer
2021-05-07T14:13:37-0400

Given statement is-

P(n):4n4n,nNP(n):4^n\ge 4n ,\forall n\in N


For n=1,LHS=41=4n=1, LHS= 4^1=4

RHS=4(1)=4=LHSRHS=4(1)=4=LHS

Hence P(n) is true for n=1.


Let us assume that p(k) is true for some positive integer k.

P(k):4k4k      (1)P(k):4^k\ge 4k~~~~~~-(1)


We want to prove p(k+1) is true when p(k) is true-


P(k+1):LHS=4k+1=4k.4P(k+1): LHS= 4^{k+1}=4^k.4

RHS=4(k+1)=4k+4RHS=4(k+1)=4k+4


As 4k44k.44k+44^k\ge 4\Rightarrow 4^k.4\ge 4k+4


So P(k+1)P(k+1) is true.


So given statement is true for all nN.n\in N.


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