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    Hi guys, just come across something in the Core Mathematics 2 textbook that I'm a bit stuck on (page 84, q6) <-- just incase I write it out horribly

    I've managed to solve this question with a little nudge in the right direction but I was wondering if there was a different, slightly quicker method to solving it. Here's the question;

    The coefficient of x^2 in the binomial expansion of (1 + \frac{x}{2})^n, where n is a positive integer, is 7. Find the value of n.

    Okay, so I figured that the coefficient of x^2, which is 7, is equal to \frac{n}{4}. From that I get 28, and that's where I got stuck. Someone told me that I know the answer is 8 because 28 is in the expansion of (a+b)^8, but surely there's a quicker way than sitting there writing out pascal's triangle all the way to 8?! :eek:

    Any help is greatly appreciated as usual :P
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    how did you get n/4
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    (Original post by lebron_23)
    Hi guys, just come across something in the Core Mathematics 2 textbook that I'm a bit stuck on (page 84, q6) <-- just incase I write it out horribly

    I've managed to solve this question with a little nudge in the right direction but I was wondering if there was a different, slightly quicker method to solving it. Here's the question;

    The coefficient of x^2 in the binomial expansion of (1 + \frac{x}{2})^n, where n is a positive integer, is 7. Find the value of n.

    Okay, so I figured that the coefficient of x^2, which is 7, is equal to \frac{n}{4}. From that I get 28, and that's where I got stuck. Someone told me that I know the answer is 8 because 28 is in the expansion of (a+b)^8, but surely there's a quicker way than sitting there writing out pascal's triangle all the way to 8?! :eek:

    Any help is greatly appreciated as usual :P
    Well:

    (1 + x/2)^n

    1 + nx + n(n-1)/2! (x/2)^2 +...

    So:

    [n(n-1)/2!]/4 = 7/4
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    Ignore my post actually.

    I'm blabbering.
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    (Original post by TenOfThem)
    how did you get n/4
    Well, I used the binomial expansion as you normally would but with the power as n rather than a number. So the expansion in ascending order of x gave me \displaystyle \binom{n}{2} x 1^n-1 x \frac{x^2}{4}. Then, making 7 = \frac{n}{4} I got 28 as my answer. If I went wrong somewhere, please point it out and please feel free to give me tonnes and tonnes of help

    Thanks
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    (Original post by lebron_23)
    Well, I used the binomial expansion as you normally would but with the power as n rather than a number. So the expansion in ascending order of x gave me \displaystyle \binom{n}{2} x 1^n-1 x \frac{x^2}{4}. Then, making 7 = \frac{n}{4} I got 28 as my answer. If I went wrong somewhere, please point it out and please feel free to give me tonnes and tonnes of help

    Thanks
    You have not used the binomial expansion at all

    (1+\frac{x}{2})^n = 1 + n(\frac{x}{2}) + \frac{n(n-1)}{2!}(\frac{x}{2})^2
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    (Original post by TenOfThem)
    You have not used the binomial expansion at all

    (1+\frac{x}{2})^n = 1 + n(\frac{x}{2}) + \frac{n(n-1)}{2!}(\frac{x}{2})^2
    Oh, that's a bit of a problem then . I thought it was possible to use the nCr function to find the coefficients of the terms in the expansion? That's sort of how I got to my answer.. It seems to work, but please explain your method, it seems more expansion-ish. Thanks again lol..
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    (Original post by TenOfThem)
    You have not used the binomial expansion at all

    (1+\frac{x}{2})^n = 1 + n(\frac{x}{2}) + \frac{n(n-1)}{2!}(\frac{x}{2})^2
    That's what I tried to type up!
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    (Original post by L'Evil Fish)
    That's what I tried to type up!
    Haha its cool, I understood. What I didn't understand was what I said in my above post :confused: meh
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    (Original post by lebron_23)
    Oh, that's a bit of a problem then . I thought it was possible to use the nCr function to find the coefficients of the terms in the expansion? That's sort of how I got to my answer.. It seems to work, but please explain your method, it seems more expansion-ish. Thanks again lol..
    I have used nCr but in the simplified version that you should have also been taught

    You seemed to decide that nC2 = 1... why?
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    (Original post by TenOfThem)
    I have used nCr but in the simplified version that you should have also been taught

    You seemed to decide that nC2 = 1... why?
    I didn't intend to make is equal 1. I thought that nC2 was equal to 28, because that divided by 4 would give the coefficient of x^2.
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    7 = \frac{n}{4} I got 28 as my answer.
    nC2 = 28

    so, rewrite this using the formula to convert from nCr into factorials:

    n! / (n-2)!2! = 28

    then you can sort this out by removing the factorials and you should get a quadratic
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    (Original post by lebron_23)
    I didn't intend to make is equal 1. I thought that nC2 was equal to 28, because that divided by 4 would give the coefficient of x^2.
    It is
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    (Original post by TenOfThem)
    It is
    Okay, that makes me feel much better, thank you. But now the problem is, I'm not sure how to get from nC2 = 28 to what n is equal to. I figured it was 8 because 28 is in pascal's triangle under the expansion of (a+b)^8

    Oh god, scrap all of that - I just realised how to get from nC2 = 28 to n = 8. Thank you for your help and patience though!!
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    (Original post by lebron_23)
    Okay, that makes me feel much better, thank you. But now the problem is, I'm not sure how to get from nC2 = 28 to what n is equal to. I figured it was 8 because 28 is in pascal's triangle under the expansion of (a+b)^8
    By using the fact that ^nC_2 = \dfrac{n(n-1)}{2!}
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    (Original post by lebron_23)

    Oh god, scrap all of that - I just realised what I was doing wrong. Thank you for your help and patience though!!
    ok
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    (Original post by Lunch_Box)
    nC2 = 28

    so, rewrite this using the formula to convert from nCr into factorials:

    n! / (n-2)!2! = 28

    then you can sort this out by removing the factorials and you should get a quadratic
    Thank you SO much, I knew it wasn't as long and as difficult as I though it was - I just forgot about using the nCr to factorial equation *major facepalm*

    Much appreciated!!
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    (Original post by TenOfThem)
    By using the fact that ^nC_2 = \dfrac{n(n-1)}{2!}
    Haha the answer emerges.. Thanks a bunch again. This has been more informative than 99% of my lessons put together >.<
 
 
 
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