Use induction to prove that for all integers n >= 1, (2n)! < 2^n( n!)^2?

Answer 1

LHS = (2n)!
RHS = 2^n( n!)^2

for n=1,
(2n)! = (2)! = 1.2 = 2
2^n( n!)^2 = 2^1( 1!)^2 = 2
So LHS = RHS, condition is not satisfied.

for n = 2,
(2n)! = (4)! = 1.2.3.4 = 24
2^n( n!)^2 = 2^2( 2!)^2 = 16
So LHS > RHS

Your problem has errors!!!

I'll try to prove:
(2n)! >= 2^n( n!)^2 .. .... (ii)

for n = 1,

(2n)! = (2)! = 1.2 = 2
2^n( n!)^2 = 2^1( 1!)^2 = 2
So LHS = RHS, condition is satisfied.

let, for any arbitrary integer m, (ii) is satisfied.

So,
(2m)! >= 2^m(m!)^2

now, if n = m+1, (ii) becomes,
(2(m+1))! >=2^(m+1) ((m+1)!)^2

now,
(2(m+1))! = (2m+2))! = (2m+2)(2m+1) (2m)!
= 2(m+1)(2m+1) (2m)!

2^(m+1) ((m+1)!)^2 = 2.2^m (m+1)^2 (m!)^2


we have assumed, (2m)! >= 2^m(m!)^2
now, multiplying by 2(m+1)(2m+1) we get,
2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(2m+1) (m!)^2

now, 2^m 2(m+1)(2m+1) (m!)^2 = 2.2^m (m+1)(2m+1) (m!)^2

again 2m+1 > m+1

so, 2(m+1)(2m+1) (2m)! >= 2^m 2(m+1)(m+1) (m!)^2

or,
(2(m+1))! >=2^(m+1) ((m+1)!)^2

(ii) is true for n = m+1
(Proved)

Answer 2

2!*n! < 2^n*n!*n!
2<2^n*n!

Proof it for n=1 and then n = n+1

you sure you aren't missing an equal sign in the inequality?