# Prove that n! > 2n-1 , n >2 using the principle of mathematical induction.

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### 1 Answer

n! > 2n-1

Since n>1 lets consider n =2 as the first term.

When n=3

3! = 6

2*3-1 = 5

6>5

So when n = 2 the result is true.

Lets assume n = p where p>2 satisfies the inequality.

p!>2p-1 ----(1)

When n = p+1

We have to show that;

(p+1)!>2(p+1)-1

Let (1)*(p+1)

p!(p+1)>(2p-1)(p+1) ----(2)

2(p+1)-1-(2p-1)(p+1)

= 2p+2-1-2p^2-p+1

= 2+p-p^2

= 2+p(1-p)

p>2

-p<-2

1-p<-2+1

1-p<-1

p(1-p)<-p

2+p(1-p)<-p+2

since p>2 then -p+2<0

2+p(1-p)<0

Therefore;

2(p+1)-1-(2p-1)(p+1)<0

2(p+1)-1 < (2p-1)(p+1)

From (2)

p!(p+1)>(2p-1)(p+1)

(p+1)!>(2p-1)(p+1)> 2(p+1)-1

(p+1)!>2(p+1)-1

So n=p+1 the result is true.

**So from mathematical induction for n>2;**

**n! > 2n-1**

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