Binomials and ProductsDate: 6/21/96 at 14:0:30 From: Anonymous Subject: Binomials and Products In Brazil, we call a Binomial of Newton a number that is represented by /n\ , and that is equal to n!/k!*(n-k)!. \k/ If S = 1*2*3 + 2*3*4 + 3*4*5 + 4*5*6+...+48*49*50, how can I represent the answer of S by using a Binomial of Newton? I have already noticed that the first term (1*2*3) is equal to 3!/0!, and that, therefore, S can be also represented by S=3!/0! + 4!/1! + 5!/2!+...+50!/47!, in an order that resembles n!/k!*(n-k)!, but that really is n!/(n-k)!. A colleague of mine has told me that I could solve this problem by trying to relate it to the columns of the Pascal's Triangle, and I think he is right, although I cannot figure it out. Could you help me? P.S. I hope my explanation is not confusing, and sorry if it is. I am using / \ as a sign for parentheses. \ / Date: 6/21/96 at 15:50:28 From: Doctor Anthony Subject: Re: Binomials and Products You can write the series as: S = 3! + 4!/1! + 5!/2! + 6!/3! + 7!/4! + .... + 50!/47! = 3!{1 + 4!/(3!.1!) + 5!/(3!.2!) + 6!/(3!.3!) + .. + 50!/(3!.47!)} = 3!{1 + 4C3 + 5C3 + 6C3 + ...+ 50C3} = 3!{4C4 + 4C3 + 5C3 + 6C3 + ... + 50C3} Now use the fact that nCr + nC(r+1) = (n+1)C(r+1) = 3!{(4C4+4C3) + 5C3 + 6C3 + . + 50C3} = 3!{(5C4 + 5C3) + 6C3 + ... + 50C3} = 3!{(6C4 + 6C3) + 7C3 + ... + 50C3} We continue this process combining pairs of terms as shown until we arrive at: S = 3!{50C4 + 50C3} S = 3!*51C4 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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