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Question

Prove that (2n)!22n(n!)213n+1 for all nϵN.

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Solution

P(n) : (2n)!22n(n!)213n+1

For n = 1

2!22.114

=1212

P(n) is true for n = 1

Let P(n) is true for n = k, so

(2n)!22n(n!)213n+1 .....(1)

We have to show that,

2(k+1)!22(k+1)[(k+1)!]213k+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).13k+1[Using equation (i)]=(2k+1)2(k+1).13k+1=(2k+2)2(k+1).13k+3+1=.13k+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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