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Question

Prove 2n!(n!)2>4n2n+1

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Solution

we can verify for n = 2 as in part (i)
Assume p(n) i. e.
(2n)!(n!)2>4n2n+1
p(n + 1) = (2n+2)![(n+1)!]2
=(2n+2)(2n+1)(n+1)2(2n)!(n!)2
=2(2n+1)n+1p(n)>2(2n+1)n+14n2n+1
=8nn+1
We have prove p( n +1) > 4(n+1)2n+3
or 8nn+1>4(n+1)2n+1
or 2n(2n+3) (n2 +2n+1)>0
or 3n2 +3n+n1>0
or 3n(n+1)+(n1)>0
Above is true for n > 1

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