Prove that 2n≥2n,∀n≥1
For any integer n ≥ 1, let Pn be the statement that
2n≥2n Base Case. The statement P1 says that
21≥2⋅1 which is true.
Inductive Step. Fix k≥1, and suppose that Pk holds, that is,
2k≥2k It remains to show that Pk+1 holds, that is,
2k+1≥2(k+1)
2k≥2,∀k≥12k≥2k=>2⋅2k≥2⋅2k=>=>2k+1≥4k=2k+2k≥2k+2=2(k+1) Therefore Pk+1 holds.
Thus, by the principle of mathematical induction, for all n≥1,Pn holds.
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