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    Problem 509

    Find \displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}
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    Solution 509

    Let  S_n = \displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n} = \frac{1}{2} + \frac{2}{4} + \frac {3}{8} + \frac {4}{16} +...   *
    Therefore \frac{1}{2} S_n = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} +... **

    Subtracting ** from *:

     \frac{1}{2}S_n = (\frac{1}{2} + \frac{2}{4} + \frac{3}{8} +....) - (\frac{1}{4} + \frac{2}{8} + \frac{3}{16} +...)
    \implies \frac{1}{2} S_n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} +... = \frac{0.5}{1-0.5} = 1 \implies S_n = 2
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    (Original post by 16Characters....)
    Solution 509
    Very nice!

    There's a neat trick to it at as well,
    Spoiler:
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    Differentiate the geometric sum formula
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    (Original post by Zacken)
    Very nice!

    There's a neat trick to it at as well,
    Spoiler:
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    Differentiate the geometric sum formula
    Nice. I'll finish my current graphs question and then try and think up some not involving matrices.
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    (Original post by 16Characters....)
    Nice. I'll finish my current graphs question and then try and think up some not involving matrices.
    I need to go practice some STEP, III 2000 is calling out to me. :rofl:
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    (Original post by Renzhi10122)
    Problem 509

    Let  a,b,c be the sides of a triangle. Prove that
     \dfrac{a}{b+c}+\dfrac{b}{c+a}+ \dfrac{c}{a+b}+\dfrac{3abc}{(a+b  )(b+c)(c+a)}<2
    Spoiler:
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    This is equivalent to (eg via algebra bash ie expand everything out)
     \displaystyle (a+b-c)(b+c-a)(c+a-b) > 0
    which is true by the triangle inequality, assuming non-degeneracy.
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    (Original post by metaltron)
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    This is equivalent to (eg via algebra bash ie expand everything out)
     \displaystyle (a+b-c)(b+c-a)(c+a-b) > 0
    which is true by the triangle inequality, assuming non-degeneracy.
    Yep, quite nice I think.
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    (Original post by Zacken)
    Problem 503

    Given integers a and b satisfying \dfrac{a}{b}  < 1, find integers c and d such that

    \displaystyle \frac{a}{b} + \frac{c}{d} + \frac{ac}{bd} = 1
    Solution 503

    Multiply out by bd.

     ad+bc+ac=bd
     (b-a)(c+d)=2bc
    We may assume that  (a,b)=1 so  c|b-a . Likewise,  (c,d)=1 so  b|c+d . 2 divides exactly one of b-a and c+d. Thus, we get two solution for c and d:
     c=b-a, d=b+a and
     c=\dfrac{b-a}{2} , d=\dfrac{b+a}{2} and these do indeed work
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    (Original post by Renzhi10122)
    Solution 503

    Multiply out by bd.

     ad+bc+ac=bd
     (b-a)(c+d)=2bc
    We may assume that  (a,b)=1 so  c|b-a . Likewise,  (c,d)=1 so  b|c+d . 2 divides exactly one of b-a and c+d. Thus, we get two solution for c and d:
     c=b-a, d=b+a and
     c=\dfrac{b-a}{2} , d=\dfrac{b+a}{2} and these do indeed work
    I agree with you up to the before last line, love the solution although it was a little overkill!

    If we have a = 4, b=5 \implies \dfrac{a}{b} = \dfrac{4}{5}

    So, c = b-a = 1 and d = 5+4 = 9 definitely works, I agree.

    But c = \dfrac{5-4}{2} = \dfrac{1}{2} \notin \mathbb{Z}
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    (Original post by Zacken)
    I agree with you up to the before last line, love the solution although it was a little overkill!

    If we have a = 4, b=5 \implies \dfrac{a}{b} = \dfrac{4}{5}

    So, c = b-a = 1 and d = 5+4 = 9 definitely works, I agree.

    But c = \dfrac{5-4}{2} = \dfrac{1}{2} \notin \mathbb{Z}
    Very good point, ignore the last line
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    Problem 510*

    Let M be a point inside a triangle ABC. Prove that:

    AB+BC+CA>MA+MB+MC+min(MA,MB,MC)
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    (Original post by Renzhi10122)
    Problem 510*

    Let M be a point inside a triangle. Prove that:

    AB+BC+CA>MA+MB+MC+min(MA,MB,MC)
    Presumaby A, B, C are the vertices of the triangle?
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    (Original post by 16Characters....)
    Presumaby A, B, C are the vertices of the triangle?
    Woops, yes they are, let me edit that...
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    (Original post by Zacken)
    Problem 509

    Find \displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}
    \begin{aligned} \displaystyle \bigg(\sum_{n \ge 0}q^n\bigg)^2 & = \bigg(\sum_{n \ge 0}q^n\bigg) \bigg(\sum_{n \ge 0}q^n\bigg) = \sum_{n \ge 0} \sum_{ 0 \le k \le n} q^k q^{n-k} \\& = \sum_{n \ge 0} \sum_{ 0 \le k \le n} q^n = \sum_{n \ge 0}(n+1)q^n. \end{aligned}

    Therefore \displaystyle \bigg(\sum_{n \ge 0}q^n\bigg)^2-\sum_{n \ge 0}q^n = \sum_{n \ge 0}nq^n

    Set q = 1/2 we have the required sum as 2^2-2 = 2.
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    (Original post by Kummer)
    x
    Nice! This is a pretty involved solution, you're pretty good at this?

    Are you at uni?
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    (Original post by Zacken)
    Nice! This is a pretty involved solution, you're pretty good at this?

    Are you at uni?
    Not that anyone would need this technique for this sum, but I think different solutions are good. Yes, I'm at uni.
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    (Original post by Zacken)
    Problem 509

    Find \displaystyle \sum_{n=0}^{\infty} \frac{n}{2^n}
    Seems like a binomial expansion will work here:

     4=(1-\frac{1}{2})^{-2}=1+\frac{2}{2}+\frac{3}{4}+ \frac{4}{8}...=2S_n
    It follows that  S_n=2
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    (Original post by Renzhi10122)
    Seems like a binomial expansion will work here:

     4=(1-\frac{1}{2})^{-2}=1+\frac{2}{2}+\frac{3}{4}+ \frac{4}{8}...=2S_n
    It follows that  S_n=2
    I think you can just look at Sn and 0.5Sn, take one from the other, and be left with a geometric series... though I haven't put pen to paper with this one.

    Posted from TSR Mobile
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    (Original post by Krollo)
    I think you can just look at Sn and 0.5Sn, take one from the other, and be left with a geometric series... though I haven't put pen to paper with this one.

    Posted from TSR Mobile
    Yep, I saw, just thought I'd give my solution.
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    (Original post by Renzhi10122)
    Yep, I saw, just thought I'd give my solution.
    Good good. Always nice to have a good variety :ahee:

    Posted from TSR Mobile
 
 
 
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