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 January 4th, 2012, 03:59 AM #1 Newbie   Joined: Jun 2011 Posts: 17 Thanks: 0 Permutation and Combination, Binomial Theorem and Hyperbola Hi First of all, i am sorry that i have posted the questions of three different topics in a same thread. But i have to do it as it will take a lot of time writing them and less functions are available on the forum editor.So i am posting image of 12 questions.
 January 4th, 2012, 08:37 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,144 Thanks: 1003 Re: Permutation and Combination, Binomial Theorem and Hyperb What is YOUR question? Is this homework? Surely you'll understand that since you didn't show your work, we can't tell where you're stuck.
 January 4th, 2012, 10:12 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Permutation and Combination, Binomial Theorem and Hyperb Are these questions part of an ongoing competition? We generally ask that only a few questions be posted per topic. (A search of your posts shows I have told you this before...) All of these questions can be written using $\LaTeX$. If the deadline has passed for this test, and you are merely hoping to find how to get the answers for your own edification, then let us know, and also what you have done and where you are stuck.
 January 5th, 2012, 04:25 AM #4 Newbie   Joined: Jun 2011 Posts: 17 Thanks: 0 Re: Permutation and Combination, Binomial Theorem and Hyperb Are you saying about this http://www.tug.org/protext/. It says its about 750 MB in size. It will be too large for downloading and installing on a cyber-cafe machine. I just don't get to know how to start the problems. I don't know why i am able to solve M.L. Khanna book's questions but why not these. I need a Key Idea here and yes the test date has expired. Thanks
 January 5th, 2012, 01:00 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Permutation and Combination, Binomial Theorem and Hyperb 3.) The binomial theorem gives us: $(1+x)^{17}=\sum_{k=0}^{17}{17 \choose k}x^{k}$ We are told: ${17 \choose 2r}>\frac{1}{2}{17 \choose 2r+1}$ $\frac{17!}{(2r)!(17-2r)!}>\frac{17!}{2(2r+1)!(17-(2r+1))!}$ $(2r)!(17-2r)(16-2r)!<2(2r+1)(2r)!(16-2r)!$ $17-2r<4r+2$ $15<6r$ We also know $2r+1\le17$ or $r\le8$ thus r can be 3,4,5,6,7,8 or 6 possible values. 4.) $$$\frac{x}{\(x^{\frac{1}{3}}-x^{\frac{7}{3}}$$^3}\)$$\frac{x^4}{\(1+x^{-2}$$^3}\)=$$\frac{1}{\(1-x^2$$^3}\)$$\frac{x^{10}}{\(1+x^2$$^3}\)=\frac{x^{ 10}}{$$1-x^4$$^3}$ The series expansion at x = 0 is: $\sum_{k=1}^{\infty}\frac{k(k+1)}{2}x^{10+4(k-1)}$ where the first term is $x^{10}$
 January 6th, 2012, 03:28 AM #6 Newbie   Joined: Jun 2011 Posts: 17 Thanks: 0 Re: Permutation and Combination, Binomial Theorem and Hyperb Thanks @MarkFL I got up to last step of I am not able to understand what you have written after that. Can i understand like this? We have got a ${x^{10}$ on the numerator. If we want to coefficient of ${x^{10}$, the binomial series which is in the denominator must be equal to one because any power of x other than ${x^{0}$ comes in expansion of $(1-x^4\)^3$will decrease ${x^{10}$ on the denominator. Hence the denominator must be equal to one which is get by putting x=0 in the denominator.Hence the answer is 1 Am i right? Thanks

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