Answer to Question #94861 in Discrete Mathematics for Niklta

Question

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

Answer 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.