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Math Help - Binomial Expansion

  1. #1
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    Binomial Expansion

    Hi Guys,

    Need a little help with this one, Not sure how to proceed.

    Two positive integers, a,b are connected by the relation a = b - 1. By using a binomial expansion show that for all positive integral values of n the expression a^{2n} + 2nb - 1 is exactly divisible by b^2. By choosing suitable values of a and n, show that 2^{40} + 119 is divisible by 9, and hence that 2^{39} + 1 is divisible by 9.

    I tried putting the substitution for a into the equation, and did a binomial expansion of (b - 1)^{2n}, but not sure where that is supposed to lead. It seems like another longer equation.

    Thanks for your help.
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  2. #2
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    Quote Originally Posted by mathguy80 View Post
    Hi Guys,

    Need a little help with this one, Not sure how to proceed.

    Two positive integers, a,b are connected by the relation a = b - 1. By using a binomial expansion show that for all positive integral values of n the expression a^{2n} + 2nb - 1 is exactly divisible by b^2. By choosing suitable values of a and n, show that 2^{40} + 119 is divisible by 9, and hence that 2^{39} + 1 is divisible by 9.

    I tried putting the substitution for a into the equation, and did a binomial expansion of (b - 1)^{2n}, but not sure where that is supposed to lead. It seems like another longer equation.

    Thanks for your help.
    I suggest you write out the first few terms and the last few terms of (b - 1)^{2n}. Then add 2nb and subtract 1. You can't help but notice something ....
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  3. #3
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    Hello,

    Some steps have been volutarily skipped.

    let's expand



    Considering it mod bē, we can remove the powers of b superior to 2, because if for example , b^2 can be factored out and since b^2=0 mod b^2, it cancels all the terms with k>2.

    Hence

    and it gives your result
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  4. #4
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    The Binomial Theorem gives (b-1)^{2n}=1-\binom{2n}{1}b+\binom{2n}{2}b^2+\cdots. Note that \binom{2n}{1}=2n and that all terms after the second are divisible by b^2. Thus, (b-1)^{2n}=1-2nb+Ab^2, where A is some integer.

    Therefore, a^2+2nb-1=Ab^2.

    This is essentially the same idea as in Moo's post.
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  5. #5
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    Thanks guys. @Moo, sorry to sound stupid. I got the summation part, but the rest of the notation went way over my head.

    @melese. Got it! The 1 and 2nb terms cancel out, rest all terms have powers of b >= 2. Then rest a=2,b=3 and n=20, forms similar expression hence divisible, which checks out. Thanks!
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  6. #6
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    What part exactly ? C_{2n^k} corresponds to
    And the sum goes to 1 because after 2 it has no interest (=0)
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  7. #7
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    Yup, I was confused by the C_ notation. So that corresponds to n choose k. That clears things up, thanks!
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