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Math Help - Binomial expansion

  1. #1
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    Binomial expansion

    I've been given the first three terms in the expansion of (1+ay)^n:

    1, 12y, and 68y^2.

    I'm asked to evaluate a and n using the fact that:
    (1+ay)^n = 1 + nay + n(n-1)/2(ay)^2 +.....

    Any help in this problem would be appreciated, what is meant by evaluate?
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  2. #2
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    Hello jake.davis
    Quote Originally Posted by jake.davis View Post
    I've been given the first three terms in the expansion of (1+ay)^n:

    1, 12y, and 68y^2.

    I'm asked to evaluate a and n using the fact that:
    (1+ay)^n = 1 + nay + n(n-1)/2(ay)^2 +.....

    Any help in this problem would be appreciated, what is meant by evaluate?
    Evaluate means 'find the value of ...'

    So, by comparing the coefficients of y and y^2, we have to find the values of a and n, for which:

    12 = na

    and 68 = \frac{n(n-1)}{2}\cdot a^2

    Square the first equation:

    144 = n^2a^2

    If we divide the second equation by this one:

    \frac{68}{144}= \frac{n-1}{2n}

    Can you continue?

    Grandad
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  3. #3
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    Upon continuing I determined a=0, n=12. Does that seem right?
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  4. #4
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    Quote Originally Posted by jake.davis View Post
    Upon continuing I determined a=0, n=12. Does that seem right?
    Starting with equation I gave you:

    \frac{68}{144}=\frac{n-1}{2n}

    \Rightarrow \frac{17}{36}=\frac{n-1}{2n}

    \Rightarrow 34n = 36(n-1) = 36n - 36

    \Rightarrow n = 18

    \Rightarrow 18a = 12

    \Rightarrow a = \tfrac23

    Grandad
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  5. #5
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    I fully understand now Grandad, thank you very much for your patience.
    I was wondering how you knew to square the first equation, I have never been taught that.

    Thanks again

    Jake
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