Question #115558

Prove that n ! > 2^n for n a positive integer greater than or equal to 4.

Prove that LHS = RHS

Expert's answer

We apply the method of mathematical induction.

Induction basis : n=4n=4



{n!=4!=1234=242n=24=164!=24>16=24\left\{\begin{array}{l} n!=4!=1\cdot2\cdot3\cdot4=24\\ 2^n=2^4=16 \end{array}\right.\longrightarrow 4!=24>16=2^4

Induction assumption : suppose that the inequalities hold for all values knk\le n



k!>2k,k=1,,nk!>2^k, k=\overline{1,\ldots,n}

Induction step : need to prove that



(n+1)!>2n+1\left(n+1\right)!>2^{n+1}

Proof.



(n+1)!=(n!)(n+1)>2n(n+1)2n(n+1)>2n+1÷(2n)n+1>2n>1true inequality, since by hypothesisn>4\left(n+1\right)!=\left(n!\right)\cdot\left(n+1\right)>2^n\cdot\left(n+1\right)\to\\[0.3cm] \left.2^n\cdot\left(n+1\right)>2^{n+1}\right|\div\left(2^n\right)\\[0.3cm] n+1>2\to\\[0.3cm] n>1-\text{true inequality, since by hypothesis}\,\,\,n>4

Q.E.D.

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