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Math Help - Maths Problems

  1. #1
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    Exclamation Maths Problems

    a) Is it possible to arrange the integers from 1 to 1000 in a table with 20 rows and 50 columns so that if for each of the rows all the numbers in it are added together, then twenty consecutive integers are obtained?

    b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by xwrathbringerx View Post
    b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
    \sum_{n=1}^{667} \frac{1}{(3n-1)(3n+2)} =
    \sum_{n=1}^{667} \frac{1}{9n^2+3n-2}
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  3. #3
    Moo
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    Hello,

    You will need this formula : 1+2+...+n=n(n+1)/2
    Quote Originally Posted by xwrathbringerx View Post
    a) Is it possible to arrange the integers from 1 to 1000 in a table with 20 rows and 50 columns so that if for each of the rows all the numbers in it are added together, then twenty consecutive integers are obtained?
    Let n, n+1, ..., n+19 be these 20 consecutive integers.
    The sum of n, n+1, ..., n+19 is also the sum of the integers from 1 to 1000.

    n+(n+1)+...+(n+19)=20n+(1+2+...+19)=20n+20*19/2=20n+190
    1+...+1000=1000*1001/2=500500

    So find n such that 20n+190=500500
    2n=50050-19, which is odd, whereas 2n is obviously even. Thus it is not possible to find an integer n satisfying the conditions you want.

    Therefore it is not possible to draw such a table.

    b) Simplify: 1/(2*5) + 1/(5*8) + 1/(8*11) + ... + 1/(2000*2003)
    Note that 5=2+3, 8=2+3*2, etc... and 2003=2+3*667

    Thus the sum is
    Maths Problems-1.gif

    Let's find a telescoping sum (in most of this type of exercises, it's the trick).
    Find a and b such that
    Maths Problems-2.gif

    Get the common denominator of the RHS :

    Maths Problems-3.gif

    Therefore 3a+3b=0 and 5a+2b=1
    --> b=-a --> 5a-2a=1 --> a=1/3 and b=-1/3

    Thus this sum is equal to


    Maths Problems-4.gif

    And then
    Maths Problems-5.gif


    \frac{1}{(2+3k)\cdot (2+3(k+1))}=\frac{a}{2+3k}+\frac{b}{2+3(k+1)}

    \begin{aligned} \frac{1}{(2+3k) \cdot (2+3(k+1))}=\frac{a}{2+3k}+\frac{b}{2+3(k+1)} &=& \frac{a(2+3(k+1))+b(2+3k)}{(2+3k) \cdot (2+3(k+1))} \\ \\
    &=& \frac{2a+3ak+3a+2b+3kb}{(2+3k) \cdot (2+3(k+1))} \\ \\
    &=& \frac{k(3a+3b)+5a+2b}{(2+3k) \cdot (2+3(k+1))} \end{aligned}

    \frac 13 \sum_{k=0}^{666} \frac{1}{2+3k}-\frac{1}{2+3(k+1)}=\frac 13 \left(\frac 12-\frac 15+\frac 15-\frac 18+\frac 18-\dots-\frac{1}{2000}+\frac{1}{2000}-\frac{1}{2003} \right)

    S=\frac 13 \left(\frac 12-\frac{1}{2003}\right)=\frac 13 \left(\frac{2001}{4006}\right)=\frac{667}{4006}
    Last edited by Moo; July 28th 2008 at 01:27 AM.
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