Question

# Prove that (2n)!22n(n!)2≤1√3n+1 for all nϵN.

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Solution

## P(n) : (2n)!22n(n!)2≤1√3n+1 For n = 1 2!22.1≤1√4 =12≤12 ⇒ P(n) is true for n = 1 Let P(n) is true for n = k, so (2n)!22n(n!)2≤1√3n+1 .....(1) We have to show that, 2(k+1)!22(k+1)[(k+1)!]2≤1√3k+4Now,=2(k+1)!22(k+1)[(k+1)!]2=2(k+1)!22k.22(k+1)!(k+1)!=(2k+2)(2k+1)(2k)!4.22(k+1)(k!)(k+1)(k!)=2(k+1)(2k+1)(2k)!4.(k+1)2.22k.(k!)2=2(2k+1)4(k+1).1√3k+1[Using equation (i)]=(2k+1)2(k+1).1√3k+1=(2k+2)2(k+1).1√3k+3+1=.1√3k+4 [Since 2k + 1 < 2k + 2 and 3k + 1 ≤ 3k + 4] ⇒ P(n) is true for n = k + 1 ⇒ P(n) is true for all nϵN by PMI.

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