Is (2k)! (2k+1) = [2(k+1)]!?
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Is (2k)! (2k+1) = [2(k+1)]!?
No it is not: http://quicklatex.com/cache3/ql_c1cd...fb17231_l3.png.Quote:
Originally Posted by dwsmith
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)!