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    I could imagine induction being used to show that the series does in fact converge to what we claim it does...
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    (Original post by rnd)
    My way was as follows.
    Spoiler:
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    S=1+4x+9x^2+.....

    Sx=x+4x^2+9x^3....

    S-Sx=1+3x+5x^2+7x^3+.......

    (S-Sx)x=x+3x^2+5x^3+.....

    S-Sx-(S-Sx)x=1+2x+2x^2+2x^3+.....

    S(1-x)^2=(1+x)/(1-x)

    S=(1+x)/(1-x)^3

    I'm sure I've seen something like this before, otherwise I probably wouldn't have solved it.

    Please, how do you get the second from last line?
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    what a piece of work is a man!

    S-Sx-(S-Sx)x=1+2x+2x^2+2x^3+.....

    so expand out brackets:

    S-Sx-Sx-Sx^2=1+2x+2x^2+2x^3+.....

    or alternatively:

    S(1-2x-x^2)=1+2x+2x^2+2x^3+.....

    and spotting the factorisation gives
    S(1-x)^2=(1+x)/(1-x)
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    (Original post by hopinmad)
    Please, how do you get the second from last line?
    i don't quite see your problem. the LHS factorises nicely, on the RHS look at it as 1+2x(1+x+x^2+...) sum the series and then fiddle around with fractions.
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    (Original post by Totally Tom)
    You're such an Andrew Ewart it's unbelievable.
    I don't know who he is.

    (Original post by tommm)
    Fourier series? From the basic grounding I have in Fourier series, I can see no connection whatsoever between them and this question.

    And how would induction work when you're proving a proposition about x, which isn't necessarily integral?
    Okay, the fourier series bit is a bit of a throw out idea. I was wondering apart from calculus can somebody solve it using a more advanced method.

    I mean use induction to show it converges to what it does.
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    (Original post by Simplicity)
    I don't know who he is.


    Okay, the fourier series bit is a bit of a throw out idea. I was wondering apart from calculus can somebody solve it using a more advanced method.

    I mean use induction to show it converges to what it does.
    he goes by the alter ego 'question boy'

    he's in the year above me and tries to prove everything by induction.

    and he hasn't accepted my friend request on fb yet >:[
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    (Original post by Totally Tom)
    i don't quite see your problem. the LHS factorises nicely, on the RHS look at it as 1+2x(1+x+x^2+...) sum the series and then fiddle around with fractions.

    Thanks, I see now.
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    (Original post by rnd)
    My way was as follows.
    Spoiler:
    Show

    S=1+4x+9x^2+.....

    Sx=x+4x^2+9x^3....

    S-Sx=1+3x+5x^2+7x^3+.......

    (S-Sx)x=x+3x^2+5x^3+.....

    S-Sx-(S-Sx)x=1+2x+2x^2+2x^3+.....

    S(1-x)^2=(1+x)/(1-x)

    S=(1+x)/(1-x)^3

    I'm sure I've seen something like this before, otherwise I probably wouldn't have solved it.
    I think the method was in a STEP III (possibly II) question from a paper definitely within the last ten years. If I may say so you choose the most elegant solution by far - it is just such an unexpected and clever method :yes:
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    (Original post by DeanK22)
    I think the method was in a STEP III (possibly II) question from a paper definitely within the last ten years. If I may say so you choose the most elegant solution by far - it is just such an unexpected and clever method :yes:
    You may say so but I'm taking no credit. I've been trying to remember and I think that what I saw before was probably 1+3x+5x^2+7x^3... which meets the other problem halfway so to speak.

    Has anyone tried any of the other problems from that paper? There are a few easy ones but also some more interesting ones.
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    The next problem up in a sequence of sums is 1+5x+14x^2+30x^3+55x^4.... which is now no harder than the other.
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    Here the coefficients a(i) are defined inductively by

    a(0)=1
    a(i) = a(i-1)+(i+1)^2 for i>0

    So then we can express the sum S as

    S = a(0) + (a(0) + 2^2)x + (a(1) + 3^2)x^2 + (a(2) + 4^2)x^3 +...
    = 1 + xS + 4x + 9x^2 +16x^3 +...
    = xS + the old series

    So we can solve for S; S= (1+x)/(1-x)^4
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    I really quite liked this one that went like this:

    

1+ \sqrt{3} \qquad \text{is a root of}  \qquad 3x^3 + ax^2 + bx +12=0

    where a and b are integers. Find the values of a and b. Oviously, as they're integers, one of the other roots is \pm (1- \sqrt{3})
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    (Original post by jbeacom600)
    I really quite liked this one that went like this:

    

1+ \sqrt{3} \qquad \text{is a root of}  \qquad 3x^3 + ax^2 + bx +12=0

    where a and b are integers. Find the values of a and b. Oviously, as they're integers, one of the other roots is \pm (1- \sqrt{3})
    I did that one too.

    \pm ?
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    You could conceivably have -1 - sqrt 3...

    i think the brackets are misplaced.
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    My reasoning went like:
    

(1+\sqrt{3})(1-\sqrt{3}) = 1 -3 =-2

-(1+\sqrt{3})(1-\sqrt{3}) =2

    it wouldn't be  -1 -\sqrt{3} as that is  -(1+\sqrt{3}) so you would get -(1+\sqrt{3})^2 and this is clearly neither rational nor an element of the integers.

    I.e. we're looking at conjugate surds. As for the other question, I think I did it the same way as tommm, but using the limits on the integral. That is of course, assuming that we can actually integrate/differentiate an infinite series term by term.
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    (Original post by jbeacom600)
    As for the other question, I think I did it the same way as tommm, but using the limits on the integral.
    What's that about? I don't think I did it that way, I KNOW I did it that me am tired
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    (Original post by rnd)
    Has anyone tried any of the other problems from that paper? There are a few easy ones but also some more interesting ones.
    I did 13 using the beta and gamma functions, but it's a bit high-powered, having to at least justify two inversions. Is there a more elementary (possibly easier) way?
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    The general term in the sequence is r!/(r+5)! = 1/((r+1)(r+2)(r+3)(r+4)(r+5)).

    Maybe you can express this in partial fractions which will then cancel in the whole sum..

    Not that I've bothered to do this.
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    (Original post by Drederick Tatum)
    .
    Given exam conditions, I might actually slog through that... Maybe there's a trick from this step, though?
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    Spoiler:
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    Consider \displaystyle \frac{1}{(r+1)(r+2)(r+3)(r+4)} - \frac {1}{(r+2)(r+3)(r+4)(r+5)}
 
 
 
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