How do you simplify (2k)!/(2k+2)! to 1/(2(k+1)(2k+1)) ?

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Answer 1 · 2018-02-06T01:11:02

The expression we want to simplify is

#E = ((2k)!)/((2k+2)!)#.

Since #(2k+2)! = (2k+2)(2k+1)(2k)!#, then:

#E = ((2k)!)/((2k+2)(2k+1)(2k)!)#.

Then:

#E = cancel((2k)!)/((2k+2)(2k+1)cancel(2k)!)#.

This gives

#E = 1/((2k+2)(2k+1))#.

We can rewrite #(2k+2) = 2(k+1)#; then:

#E = 1/(2(k+1)(2k+1))#.

Hope it helped!