If 4nn+1<(2n)!(n!)2 then P(n) is true for

Let P(n):4nn+1<(2n)!(n!)2For n=2P(2):422+1<4!(2)2⇒163<244which is true.Let for n, = m≥2 , P(m) is true.i.e 4mm+1<(2m)!(m!)2now,4n+1m+2=4mm+1⋅4(m+1)m+2<(2m)!(m!)2⋅4(m+1)(m+2)=(2m)!(2m+1)(2m+2)4(m+1)(m+1)2(2m+1)(2m+2)(m!)2(m+1)2(m+2)=[2(m+1)]![(m+1)!]2⋅2(m+1)2(2m+1)(m+2)<[2(m+1)]![(m+1)!]2Hence, for n≥ 2, P(n) is true.