Perfect Square EquationDate: 02/22/2002 at 09:10:25 From: Barry Hynes Subject: Prove perfect square equation Could somebody explain the following equation for me, please? Prove that if n > 1, then nC2 + (n-1)C2 is a perfect square. For n = 2 the answer is 1; for n = 3 the answer is 3 + 1 = 4; and for n = 4 the answer is 6 + 3 = 9. How do you prove this? Through Sigma or by expanding nC2 into n(n-1)!/2!*(n-2)!... Barry Date: 02/22/2002 at 11:23:53 From: Doctor Paul Subject: Re: Prove perfect square equation My first instinct is to say that using factorials is the way to go, although as I don't know if this will actually work. Let's see what happens: C(n,2) + C(n-1,2) = n! (n-1)! --------- + --------- = 2! (n-2)! 2! (n-3)! n*(n-1)/2 + (n-1)*(n-2)/2 = n*(n-1) + (n-1)*(n-2) --------------------- = 2 (2*n^2 - 4*n + 2)/2 = n^2 - 2*n + 1 = (n-1)^2 So this seems to be the way to go. I'm sure there are other ways to prove this identity, but this one was the most obvious to me. I hope this helps. Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ Date: 02/24/2002 at 09:06:09 From: Barry Hynes Subject: Prove perfect square equation Hi, Thanks for the help. The way you solved it was the way I was trying to do it, but I had the following and cannot see how you got to n*(n-1)/2 + (n-1)*(n-2)/2 = What happened to the ! ? My math skills have evaporated over the past 15 years. MY WAY n! (n-1)! --------- + --------- = 2! (n-2)! 2! (n-3)! n(n-1)! + (n-1)(n-2)! ---------- ----------- (2!)(n-2)! (2!)(n-3)! --------------------------------------- YOUR WAY n! (n-1)! --------- + --------- = 2! (n-2)! 2! (n-3)! n*(n-1)/2 + (n-1)*(n-2)/2 = Thanks, Barry Date: 02/24/2002 at 16:25:36 From: Doctor Paul Subject: Re: Prove perfect square equation 5*4*3*2*1 Consider 5! / 3! = --------- = 5*4 3*2*1 10*9*8*7*6*5*4*3*2*1 What about 10! / 8! ? -------------------- = 10*9 8*7*6*5*4*3*2*1 We can generalize this: n * (n-1) * (n-2)! n! / (n-2)! = ------------------ = n * (n-1) (n-2)! Does this help? Please write back if you'd like to talk about this more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
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