Is (2k)! (2k+1) = [2(k+1)]!?
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Originally Posted by dwsmith Is (2k)! (2k+1) = [2(k+1)]!? No it is not: .
No, it would be (2k+1)! If you imagine it in reverse (2k+1)! = (2k+1) * 2k * (2k-1) * .... * 2 * 1 = (2k+1)*(2k)! On [2(k+1)]! you can do the distributive law first: (2k+2)! = (2k+2) * (2k+1) * (2k)!