My Math Forum Binomial Expansion - The notation

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 August 17th, 2018, 11:52 PM #1 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Binomial Expansion - The notation Q1: nCr+1 - nCr-1 = n+1Cr-1 - n+1Cr Q2: nC3 = 5nC1 Q3: nC1+ nC2 = 3n I would like to know how to solve these questions with detail steps, thank you!! My answer on Q2 is 1 and Q3 is 4 but the model answer are 7 and 5 respectively.
 August 18th, 2018, 06:43 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,957 Thanks: 1844 (1) Is the equation typed correctly? It reduces eventually to n = 2r - 1, where r > 1. (2) n(n - 1)(n - 2)/6 = 5n, so n((n - 1)(n - 2) - 30) = 0. Hence n = 0 or 7. (3) 3n = nC1 + nC2 = n + n(n - 1)/2, so 0 = n(2 + n - 1 - 6). Hence n = 0 or 5. Thanks from Sebastian Garth, topsquark and hy2000
 August 18th, 2018, 07:11 AM #3 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Question 1 should be Prove LHS=RHS
 August 18th, 2018, 01:36 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,957 Thanks: 1844 There must be an error in question (1), as the equation given isn't true in general. Maybe its right-hand side should have had n+1Cr+1 instead of n+1Cr-1. It then follows from nCr-1 + nCr = n+1Cr and nCr + nCr+1 = n+1Cr+1.
 August 18th, 2018, 06:59 PM #5 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Is it possible that you can prove question 1 has an error in detail steps, cause this question is from my textbook so I want to know what is wrong, thank you.
 August 18th, 2018, 07:13 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,957 Thanks: 1844 (1) In its original form, the equation doesn't hold for n = 4 and r = 2, say, so if the third term is n+1Cr-1 in the textbook, the textbook is in error.
 August 18th, 2018, 07:23 PM #7 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Would you prove that nC(r+1) - nC(r-1) = (n+1)C(r+1) - (n+1)Cr Later I contact the publisher to see if there are any typo problem in this question, thanks a lot
August 18th, 2018, 09:55 PM   #8
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Quote:
 Originally Posted by hy2000 Would you prove that nC(r+1) - nC(r-1) = (n+1)C(r+1) - (n+1)Cr Later I contact the publisher to see if there are any typo problem in this question, thanks a lot
Can you at least give it a try yourself and show us what you've done?

-Dan

August 19th, 2018, 03:17 AM   #9
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Quote:
 Originally Posted by hy2000 Would you prove that nC(r+1) - nC(r-1) = (n+1)C(r+1) - (n+1)Cr
I've already given two equations that are equivalent to Pascal's rule (also known as Pascal's identity), and which you can use without proof (but several proofs are given in the linked article). If you subtract one of the equations from the other, thus eliminating the nCr terms, the resulting equation can be rearranged to give the equation quoted above that your textbook probably intended.

Last edited by skipjack; August 19th, 2018 at 08:53 PM.

 August 19th, 2018, 04:15 PM #10 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Sure I will try it myself, if I have any question I will ask you

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