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    Got a maths C2 exam on Monday and noticed something I hadn't been taught.

    Reducing (a+b)^n to k(1+x)^n

    Can anyone help explain this to me. Cheers

    Also, on a past paper I could not get the following question


    (1+ax)^n = 1 - 20x + 180x^2

    the answer is n=10, a =-2

    Any help appreciated
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    \displaystyle \left(a+b\right)^n = \left(a\left(1+\frac{b}{a}\right  )\right)^n = a^n\left(1+\frac{b}{a}\right)^n

    For the second question, post your working.
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    (Original post by notnek)
    \displaystyle \left(a+b\right)^n = \left(a\left(1+\frac{b}{a}\right  )\right)^n = a^n\left(1+\frac{b}{a}\right)^n

    For the second question, post your working.
    Thanks.

    For the second question, I didn't have a clue. I tried to remove the (n over 1) and (n over 2) to no avail :confused:
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    From the binomial theorem:

    \displaystyle (1+ax)^n = 1+nax+\frac{n(n-1)}{2} a^2 x^2 +...

    So we have na=-20 and \frac{n(n-1)a^2}{2} = 180.

    Can you do it from here?
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    (Original post by notnek)
    From the binomial theorem:

    \displaystyle (1+ax)^n = 1+nax+\frac{n(n-1)}{2} a^2 x^2 +...

    So we have na=-20 and \frac{n(n-1)a^2}{2} = 180.

    Can you do it from here?
    Yes, I can thanks very much

    Can I ask where you got \frac{n(n-1)a^2}{2}= 180. from?
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    (Original post by the stute)
    Yes, I can thanks very much

    Can I ask where you got \frac{n(n-1)a^2}{2}= 180. from?
    The \frac{n(n-1)a^2}{2} bit comes directly from the Binomial theorem. The = 180 bit is from the series that you gave. The co-efficient of the x^2 term is 180.
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    Aah, haven't been taught that part and its not on the syllabus anymore. The question was from 1 6 year paper, thats why Im unfamiliar. Cheers guys
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    (Original post by the stute)
    Aah, haven't been taught that part and its not on the syllabus anymore. The question was from 1 6 year paper, thats why Im unfamiliar. Cheers guys
    Which exam board/exam is this? Because I thought Binomial theorem was a standard C2 question
 
 
 
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