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Math Help - Binomial expansion

  1. #1
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    Binomial expansion

    Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression p^{2n}-2nq-1 can be divided exactly by q^2 for all positive integers of n .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Two positive integers , p and q are connected by p=q+1 . By using the binomial expansion , show that the expression p^{2n}-2nq-1 can be divided exactly by q^2 for all positive integers of n .
    Using

    (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots

    we see your expression becomes

    (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots
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  3. #3
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    Quote Originally Posted by danny arrigo View Post
    Using

    (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots

    we see your expression becomes

    (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots
    Thanks Danny , but i still don see how it works . Need a little more explaination .

    Thanks again.
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    Quote Originally Posted by danny arrigo View Post
    Using

    (1+q)^{2n} = 1 + 2nq + \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots

    we see your expression becomes

    (1+q)^{2n} -2nq-1= \frac{2n(2n-1)}{2!} q^2 + \frac{2n(2n-1)(2n-2)}{3!} q^3 \cdots

    oh i see now .. thanks danny
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