Use the principle of induction Prove that 2 1 n 2 n 2 n n 1 > + ∀ > − ,

Prove that $2^n\geq2n, \forall n\geq1$

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

$2^n\geq2n$

Base Case. The statement $P_1$ says that

$2^1\geq2\cdot1$

which is true.

Inductive Step. Fix $k\geq1,$ and suppose that $P_k$ holds, that is,

$2^k\geq2k$

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

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

$2k\geq2, \forall k\geq1$$2^k\geq2k=>2\cdot2^k\geq2\cdot2k=>$$=>2^{k+1}\geq4k=2k+2k\geq2k+2=2(k+1)$

Therefore $P_{k+1}$ holds.

Thus, by the principle of mathematical induction, for all $n\geq1,P_n$ holds.