We apply the method of mathematical induction.
Induction basis : n = 4 n=4 n = 4
{ n ! = 4 ! = 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24 2 n = 2 4 = 16 ⟶ 4 ! = 24 > 16 = 2 4 \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 { n ! = 4 ! = 1 ⋅ 2 ⋅ 3 ⋅ 4 = 24 2 n = 2 4 = 16 ⟶ 4 ! = 24 > 16 = 2 4
Induction assumption : suppose that the inequalities hold for all values k ≤ n k\le n k ≤ n
k ! > 2 k , k = 1 , … , n ‾ k!>2^k, k=\overline{1,\ldots,n} k ! > 2 k , k = 1 , … , n
Induction step : need to prove that
( n + 1 ) ! > 2 n + 1 \left(n+1\right)!>2^{n+1} ( n + 1 ) ! > 2 n + 1
Proof.
( n + 1 ) ! = ( n ! ) ⋅ ( n + 1 ) > 2 n ⋅ ( n + 1 ) → 2 n ⋅ ( n + 1 ) > 2 n + 1 ∣ ÷ ( 2 n ) n + 1 > 2 → n > 1 − true inequality, since by hypothesis n > 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 ( n + 1 ) ! = ( n ! ) ⋅ ( n + 1 ) > 2 n ⋅ ( n + 1 ) → 2 n ⋅ ( n + 1 ) > 2 n + 1 ∣ ∣ ÷ ( 2 n ) n + 1 > 2 → n > 1 − true inequality, since by hypothesis n > 4
Q.E.D.
Comments
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By using mathematical induction prove that (n+1)! > 2^(n+1) for n, where n is a positive integer greater than or equal to 4