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Sigma sequence problem

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    t1, t2, t3, t4 , ...... , tn is a sequence
    t2 - t1 = t3 - t2 = t4 - t3 .............. = t(n-1) - tn

    show that:

    1/t1t2 + 1/t2t3 + 1/t3t4 ............... + 1/tn-1tn = n-1/t1tn

    2. Relevant equations



    3. The attempt at a solution
    I tried to use sigma but I couldn't solve it ..




    Please give me the beginning of the solution
     
  2. jcsd
  3. Feb 22, 2009 #2

    HallsofIvy

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    Re: Sequenec..

    Essentially this says that the sequence has a common difference and so is an "aritmetic" sequence. Let d= t2- t1= t3- t2= t4- t3 ... Then t2= t1+ d, t3= t2+ d= t1+ 2d, and, in general, tn= t1+ d(n-1).

    1/t1t2= [tex]\frac{1}{t1(t1+ d)}= \frac{A}{t1}+ \frac{B}{t1+ d}[/tex]
    What are A and B?

    1/t2t3= [tex]\frac{1}{(t1+d)(t1+2d)}= \frac{C}{t1+d}+ \frac{D}{t1+ 2d}[/tex]
    What are C and D?
     
    Last edited: Feb 22, 2009
  4. Feb 22, 2009 #3
    Re: Sequenec..

    :uhh:
    what are a, b, c, d ???
     
  5. Feb 22, 2009 #4
    Re: Sequenec..

    Look up partial fraction.
     
  6. Feb 22, 2009 #5
    Re: Sequenec..

    a = (-bt1+1)/(d+t)
    b = [ -a (d+t1) + 1 ] / t1
     
  7. Feb 22, 2009 #6

    HallsofIvy

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    Re: Sequenec..

    ??? To solve for a and b, there should be no "b" in the "a" formula or "a" in the "b" formula. Also don't use small letters for A and B. That's particularly confusing here since "d" and "D" represent different things.

    [tex]\frac{1}{t1(t1+ d)}= \frac{A}{t1}+ \frac{B}{t1+ d}[/tex]
    multiply through by the denominator t1(t1+d):
    1= A(t1+d)+ Bt1= (A+B)t1+ Ad
    so we must have A+ B= 0 and Ad= 1. A= 1/d and B= -1/d

    [tex]\frac{1}{t1t2}= \frac{1/c}\left(\frac{1}{t1}- \frac{1}{t1+d}\right)[/tex]

    For
    [tex]\frac{1}{t2t3}=\frac{1}{(t1+d)(t1+2d)}= \frac{C}{t1+d}+ \frac{D}{t1+ 2d}[/tex]
    multiply through by (t1+d)(t1+2d) to get 1= C(t1+2d)+ D(t1+ d).

    Now taking t1= -d, 1= dC, so C= 1/d and taking t1= -2d, 1= -dD, so D= -1/d.

    [tex]\frac{1}{(t1+d)(t1+2d)}= \frac{1}{d}\left(\frac{1}{t1+d}- \frac{1}{t1+ 2d}\right)[/itex]

    Putting those together,
    [tex]\frac{1}{t1t2}+ \frac{1}{t2t3} = \frac{1}{c}\left(\frac{1}{t1}- \frac{1}{t1+d}+ \frac{1}{t1+d}- \frac{1}{t1+ 2d}\right)[/itex]

    Do you see what happens? Can you do the same for the other terms?
     
    Last edited: Feb 22, 2009
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