1. ## Induction Question

It's pretty straightforward but I can't seem to get it out. Thanks in advance.

Prove by induction that for all integers n≥1: (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]

2. ## Re: Induction Question

Originally Posted by HoldingOn
It's pretty straightforward but I can't seem to get it out. Thanks in advance.

Prove by induction that for all integers n≥1: (n+1)(n+2)...(2n-1)2n= 2^n [1x3x...x(2n-1)]
For n=1: LHS (1 factor) = (1+1) = 2 = RHS

Assume true for n = k >= 1

i.e. (k+1)(k+2) . . . (2k) = 2k(1 x 3 x . . . . (2k-1)]

For n = k+1:

LHS = (k+1+1)(k+1+2)(k+1+3) . . . (2k+1)(2k+2)

= (k+2)(k+3)(k+4) . . . (2k)(2K+1) x 2 x (k+1)

= 2 x {(k+1)(k+2) . . . (2k)} x (2[k+1]-1)

= 2 x {2k (1 x 3 x . . . (2k-1)] } x (2[k+1]-1)

= 2k+1 (1 x 3 x 5 x . . . x (2[k+1] - 1)

.: if formula is true for n = k then it is also true n = k+1

.: by the principle of mathematical induction . . .

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