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    Problem 511

    Show that

    \displaystyle \sum_{n=1}^{\infty} \dfrac{n^2}{e^n} = \dfrac{e(e+1)}{(e-1)^3}
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    Solution 511
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    Let  S = \displaystyle \sum_{n=1}^{\infty} \frac{n^2}{e^n}
    Therefore  eS = \displaystyle \sum_{n=1}^{\infty} \frac{n^2}{e^{n-1}}
    Therefore  (e-1)S = \displaystyle \sum_{n=0}^{\infty} \frac{2n+1}{e^n}
     = 2 \displaystyle \sum_{n=0}^{\infty} \frac{n}{e^n} + \displaystyle \sum_{n=0}^{\infty} \frac {1}{e^n} = 2 \displaystyle \sum_{n=0}^{\infty} \frac{n}{e^n} + \frac{1}{1-\frac{1}{e}} = 2 \displaystyle \sum_{n=0}^{\infty} \frac{n}{e^n} + \frac{e}{e-1}

    Let  X = \displaystyle \sum_{n=0}^{\infty} \frac{n}{e^n}
    Therefore  eX = \displaystyle \sum_{n=0}^{\infty} \frac{n}{e^{n-1}}

    And hence  (e-1)X = 1 + \frac {1}{e} + \frac {1}{e^2} + \frac {1}{e^3} +.... = \frac{1}{1-\frac{1}{e}} = \frac {e}{e-1} \implies X = \frac{e}{(e-1)^2}


    Therefore  (e-1)S = 2X + \frac{e}{e-1} = \frac {2e}{(e-1)^2} + \frac {e}{e-1} = \frac {e(1+e)}{(e-1)^2} \implies S = \frac {e(1+e)}{(e-1)^3} .
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    (Original post by Indeterminate)
    Problem 511

    Show that

    \displaystyle \sum_{n=1}^{\infty} \dfrac{n^2}{e^n} = \dfrac{e(e+1)}{(e-1)^3}
    Solution 511

    Quite like this one.

    Let  S_n=\frac{1}{e}+\frac{4}{e^2}+ \frac{9}{e^3}+ \frac{16}{e^4}... be our sum.
    Then  S_n(1-\frac{1}{e})=\frac{1}{e}+ \frac{3}{e^2}+ \frac{5}{e^3} +...
    Then
     S_n(1-\frac{1}{e}-\frac{1}{e}(1- \frac{1}{e}))=\frac{1}{e}+ \frac{2}{e^2}+ \frac{2}{e^3}...
     S_n(\frac{e-1}{e}- \frac{e-1}{e^2})=\frac{1}{e}+ \frac{2}{e^2}(1+\frac{1}{e}+ \frac{1}{e^2}...)
     S_n(\frac{e^2-2e+1}{e^2})=\frac{1}{e}+\frac{2}  {e(e-1)}
     S_n(\frac{(e-1)^2}{e^2})=\frac{e+1}{e(e-1)}
     S_n=\dfrac{e(e+1)}{(e-1)^3}

    edit: aw, i was too slow.
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    (Original post by Renzhi10122)
    Solution 511

    Quite like this one.
    edit: aw, i was too slow.
    Haha your solution is much nicer than my "brute force" though.
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    (Original post by 16Characters....)
    Haha your solution is much nicer than my "brute force" though.
    Well, it's pretty much exactly the same tbh.
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    (Original post by Renzhi10122)
    Well, it's pretty much exactly the same tbh.
    Thought I might be the first to solve one of these for once... two people get in before me

    Anyhow my solution was similar, but naturally more messy, that the two that have gone before.

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    (Original post by Krollo)
    Thought I might be the first to solve one of these for once... two people get in before me

    Anyhow my solution was similar, but naturally more messy, that the two that have gone before.

    Posted from TSR Mobile
    My question stil remains
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    Problem 512

    Prove that in a complete bipartite planar graph connecting two sets; one of 3 elements and 1 of 2 elements, one of the nodes in the graph always lies geographically inside of the cycle formed by all of the other nodes.

    Hence deduce that the "Three Utilities Problem" is impossible. (https://en.wikipedia.org/wiki/Three_utilities_problem )
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    Nice, nice, this thread seems to be picking up, I might try crafting one or two questions later tonight if I find the time.
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    (Original post by 16Characters....)
    Problem 512

    Prove that in a complete bipartite planar graph connecting two sets; one of 3 elements and 1 of 2 elements, one of the nodes in the graph always lies geographically inside of the cycle formed by all of the other nodes.

    Hence deduce that the "Three Utilities Problem" is impossible. (https://en.wikipedia.org/wiki/Three_utilities_problem )
    Solution 512

    Consider any face. Then any two adjacent vertices will be in different sets. It follows that the face is composed of an even number of vertices, namely 4. Therefore, the outer face is comprised of 4 vertices, leaving one vertex in the area not reached by this face, i.e. the area insides the cycle made by these 4 vertices. In particular, this vertex is part of the set of 3 vertices, since the outer face is comprised of 2 vertices from each set.

    In the three utilities problem, it is clear that the outer face cannot have 6 vertices. If it did then this would lead to a contradiction immediately due to crossing lines within ths cycle formed. Therefore the outer face has 4 vertices, so 2 vertices lie inside the cycle formed. We note now that this graph can be formed by adding one vertex inside the outer cycle to the graph formed previously. However, the vertex inside this cycle splits the face into two faces, one of which our added vertex resides in. It must be joined by edges to the vertex that is in the interior of the cycle and the 2 vertices that the interoir vertex is not connected to, but this is impossible.
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    Problem 513

    Stupid Bob claimed:

    

sin(x+y) = sin(x) + sin(y)

cos(x+y) = cos(x) + cos(y)

tan(x+y) = tan(x) + tan(y)

    In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
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    (Original post by Krollo)
    Problem 513

    Stupid Bob claimed:

    

sin(x+y) = sin(x) + sin(y)

cos(x+y) = cos(x) + cos(y)

tan(x+y) = tan(x) + tan(y)

    In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
    Spoiler:
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    Would it just be something like solving \cos x = \cos y = 1 for the first one, and similarly for the rest?
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    (Original post by Krollo)
    Problem 513

    Stupid Bob claimed:

    

sin(x+y) = sin(x) + sin(y)

cos(x+y) = cos(x) + cos(y)

tan(x+y) = tan(x) + tan(y)

    In each case, find all values of x and y such that Stupid Bob is serendipitously correct.
    so for individual cases in isolation , or all cases correct simultaneously?
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    (Original post by GCSEHELPPLS)
    so for individual cases in isolation , or all cases correct simultaneously?
    In isolation, I reckon.

    Posted from TSR Mobile
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    (Original post by Krollo)
    In isolation, I reckon.

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    k thanks, I've only done the first one so far, is it worth submitting?
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    (Original post by GCSEHELPPLS)
    k thanks, I've only done the first one so far, is it worth submitting?
    Sure, go for it!
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    sin(x+y)-sin(x)-sin(y) = 0
    sin(x)cos(y)+cos(x) sin(y) - sin(x)-sin(y)=0
    sin(x) (cos(y)-1)+(cos(x)-1) sin(y) = 0

    subbing double angle formulas in for x/2 and y/2 gives
    2sin(x/2)cos(x/2)(cos^2(y/2)-sin^(y/2)-1)+2sin(y/2)cos(y/2)(cos^2(x/2)-sin^(x/2)-1)= 0

    1=sin^2(a)+cos^2(a)so we get
    2sin(x/2)cos(x/2)(-2sin^2(y) + 2sin(y/2)cos(y/2)(-2sin^2(x)=0
    -4 sin(x/2) sin(y/2) (sin(x/2)cos(y/2)+cos(x/2) sin(y/2)= 0
    -4 sin(x/2) sin(y/2) sin(x/2+y/2) = 0

    so x+y=2pi or
    sinx/2 =0 or
    siny/2=0

    for x and y being above than or equal to 0 and less than 2 pi, this gives x=2pi-y but if x or y are 0,
    the other one can be anything
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    sorry very messy - i don't know how to use latex
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    (Original post by Zacken)
    Sure, go for it!
    I've submitted in plain text, just asking how do u get latex on the post ? thanks
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    (Original post by GCSEHELPPLS)
    I've submitted in plain text, just asking how do u get latex on the post ? thanks
    Here you go. This is quite helpful:

    http://www.thestudentroom.co.uk/wiki/LaTex
 
 
 
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