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    (Original post by Mladenov)
    Hint (Problem 23)

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    Write k=n-i-j, so \displaystyle \sum_{S}ijk= \sum_{(2 \le i+j \le n-1)_{i,j \ge 1}}ij(n-i-j). Now for each m \in \{2,3,..,n-1\}, consider \displaystyle n\sum_{(i+j=m)_{i,j \ge 1}} ij, \displaystyle \sum_{(i+j=m)_{i,j \ge 1}} ji^2, and \displaystyle \sum_{(i+j=m)_{i,j \ge 1}} ij^2. Write i=m-j, and note that i takes each value from \{1,2,...,m-1\} exactly once. We are now able to evaluate each of these three sums. Then sum over \{2,3,...,n-1\} to find \displaystyle \sum_{S}ijk.


    There should be a more concise solution.
    There is

    There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods
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    Subbing post (and admission of spending a couple of hours here instead of doing revision...)
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    (Original post by shamika)
    There is

    There is a way of doing this question where I'd peg the method as **/***, but it is definitely possible with very simple methods
    It's an interesting problem, and it looks like:

    \displaystyle\sum_{(i,j,k)\in S} ijk = \dfrac{(n+2)!}{(n-3)!5!}

    and for:

    i+j+k = n+1

    x+y+z = n

    \displaystyle\sum_{(i,j,k)\in S_1} (ijk) - \displaystyle\sum_{(x,y,z)\in S_2} (xyz) = {n+3 \choose 4}

    Where {n+3 \choose 4} is the nth pentatopic number.
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    (Original post by Noble.)
    It's an interesting problem, and it looks like:

    \displaystyle\sum_{(i,j,k)\in S} ijk = \dfrac{(n+2)!}{(n-3)!5!}

    and for:

    i+j+k = n+1

    x+y+z = n

    \displaystyle\sum_{(i,j,k)\in S_1} (ijk) - \displaystyle\sum_{(x,y,z)\in S_2} (xyz) = {n+3 \choose 4}

    Where {n+3 \choose 4} is the nth pentatopic number.
    So for the original sum...
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     {n+2 \choose 5}?
    Can you prove it?
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    Solution 29

    By Van der Waerden's theorem (a_n)_{n\geq 1} must contain arbitrarily long arithmetic progressions hence there are infinitely many finite sequences (p_{n})_{1\leq n\leq k} for which a_{p_{n}}=a_{p_{1}}+(n-1)h with k>a_{p_1}. By fixing n-1=a_{p_1} we have a_{p_1}|a_{p_n} as desired. There are infinitely many such sequences hence there are infinitely many ordered pairs (p,q):\,a_{p}|a_{q}
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    Problem 30*

    More limits & integrals?

    Evaluate \displaystyle \lim_{n\to\infty} \left(\int_0^1 \frac{dx}{1+x^n}\right)^n
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    (Original post by metaltron)
    ...
    You have integrated incorrectly, \dfrac{d}{dy}e^{yn}\neq ne^{y}

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    Edit: by the way, this is going to seem pedantic but the notation \displaystyle\lim_{n\to\infty} \frac{1}{n}\to 0 is meaningless (use an equal sign)
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    (Original post by Lord of the Flies)
    You have integrated incorrectly, \dfrac{d}{dy}e^{yn}\neq ne^{y}

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    Edit: by the way, this is going to seem pedantic but the notation \displaystyle\lim_{n\to\infty} \frac{1}{n}\to 0 is meaningless (use an equal sign)
    Thanks. Got rid of it now as it was all completely wrong then! Might look at it again tomorrow when I'm more awake.

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    Thinking about it is the answer 1 as when 0 < x < 1, 1 + x^n = 1 as n tends to infinity. So the area under the curve becomes a rectangle area 1, which when raised to the power of n is still 1?
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    (Original post by Indeterminate)
    ...
    Check that expansion - it isn't valid on the interval!
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    (Original post by Lord of the Flies)
    Check that expansion - it isn't valid on the interval!

    Indeed :banghead:

    I was wondering why it seemed so easy. I'll try something different
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    Solution 30

    Using the binomial expansion and integrating, we find that:

    \displaystyle\lim_{n\to \infty} \left( \int^1_0 \frac{dx}{1+x^n} \right)^n=\lim_{n\to \infty}\left(1-\frac{1}{n+1}+\frac{1}{2n+1}-\frac{1}{3n+1}+...\right)^n\\& \displaystyle=\lim_{n\to \infty}\left(1-\left(\frac{1}{n}-\frac{1}{2n}+\frac{1}{3n}+... \right) \right)^n \\& =\displaystyle\lim_{n\to \infty}\left(1-\frac{1}{n}\int^1_0 \frac{dx}{1+x}\right)^n \\& =\displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n \\& =\displaystyle\frac{1}{2}
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    (Original post by und)
    Solution 30

    -snip-

    & =\displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n \\& =\displaystyle\frac{1}{2}
    Lol, way to trip right in front of the finish line..

    EDIT: Right, ignore this then.
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    (Original post by aznkid66)
    Lol, way to trip right in front of the finish line..
    Blithering idiot.

    \displaystyle\lim_{n\to \infty}\left(1-\frac{\ln2}{n}\right)^n & =\displaystyle\frac{1}{2}

    Lol, I got warning points for this.
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    Problem 31*

    Find all real numbers x, y and z which satisfy the simultaneous equations x^2-4y+7=0, y^2-6z+14=0 and z^2-2x-7=0.
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    (Original post by und)
    Problem 31

    Find all real numbers x, y and z which satisfy the simultaneous equations x^2-4y+7=0, y^2-6z+14=0 and z^2-2x-7=0.
    Solution 31

    Adding the three equations:

    x^2 - 2x + y^2 - 4y + z^2 - 6z =  -14

    Completing the square:

    (x-1)^2 + (y-2)^2 + (z-3)^2 = 0

    Hence x = 1, y = 2, z = 3
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    Problem 32*

    Find all positive integers n such that 12n-119 and 75n-539 are both perfect squares.
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    (Original post by und)
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    (Original post by Upper Echelons)
    ...
    I thought und answered his own question there for a second :lol:

    Solution 32
    Solution 32
    Let 12n - 119 = a^{2} and 75n - 539 = b^{2} where a, b \in \mathbb{Z} Isolating 'n' and equating the two, we get that

    4b^{2} - 25a^{2} = 819

    =(2b-5a)(2b+5a) = 819

    Running through the cases where two integers multiply together to make 819...

    (2b-5a)(2b+5a) = 819 \cdot 1, 273 \cdot 3, 117 \cdot 7, 91 \cdot 9, 63 \cdot 13, 39 \cdot 21

    Just a horrendous case bash now ;_;

    2b \mp 5a = 819, 273, 117, 91, 63, 39

    2b \pm 5a = 1, 3, 7, 9, 13, 21

    Bearing in mind that a, b are integers, we have to find the integer solutions. This only occurs when the difference of the two factor is a multiple of 10 because it reduces down to \pm 10a = ... which only works for the pairs of (273, 3), (117, 7), (63, 13)

    Solving simultaneously, for (a,b) we get the solutions ( \pm 11, 31), (\pm 27, 69), (\pm 5, 19) If we sub these back into 12n - 119 = a^{2} and 75n -539 = b^{2} ;_;

    12n - 119 = 11^{2}, 27^{2}, 5^{2} and 75n - 539 = 31^{2}, 69^{2}, 19^{2}

    We only take the integer solutions, so we get n = 12 and n = 20

    I'm sure there's a much more elegant solution than this disgusting case bash :rolleyes:
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    Problem 33

    Determine \int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx
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    (Original post by Benjy100)
    Problem 33

    Determine \int\limits_b^a {\frac{1}{{\sqrt {(a - x)(x - b)} }}} {\rm{ }}dx
    Surely this has asymptotes in the interval you've given?
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    Solution 33

    Method 1:

    Letting x=b\cos^2 t+a\sin^2 t gives:

    \displaystyle \int_b^a \frac{dx}{\sqrt{(a-x)(x-b)}} = \int_0^{\pi/2} 2 dt =\pi

    Method 2:

    \displaystyle\int_b^a \frac{1}{\sqrt{(a-x)(x-b)}}\;dx=i\int_a^b \frac{1}{\sqrt{(a-x)(b-x)}}\;dx


    =\displaystyle -2i\int_a^b\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)\;dx


    =-2i\ln(\sqrt{a-x}+\sqrt{b-x})|_a^b=-i\ln\left(\dfrac{a-b}{b-a}\right)=\pi

    (taking the principal branch - I'm sure that can be justified)
 
 
 
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