Question #94861
Use the principle of induction Prove that 2 1 n 2 n 2
n n 1
> + ∀ >

,
1
Expert's answer
2019-09-20T11:14:23-0400

Prove that 2n2n,n12^n\geq2n, \forall n\geq1

For any integer n ≥ 1, let PnP_n be the statement that


2n2n2^n\geq2n

Base Case. The statement P1P_1 says that


21212^1\geq2\cdot1

which is true.

Inductive Step. Fix k1,k\geq1, and suppose that PkP_k holds, that is,


2k2k2^k\geq2k

It remains to show that Pk+1P_{k+1} holds, that is,


2k+12(k+1)2^{k+1}\geq2(k+1)


2k2,k12k\geq2, \forall k\geq12k2k=>22k22k=>2^k\geq2k=>2\cdot2^k\geq2\cdot2k=>=>2k+14k=2k+2k2k+2=2(k+1)=>2^{k+1}\geq4k=2k+2k\geq2k+2=2(k+1)

Therefore Pk+1P_{k+1} holds.

Thus, by the principle of mathematical induction, for all n1,Pnn\geq1,P_n holds.


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