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Series question

  1. Jul 24, 2012 #1
    xQqbe.png

    I don't understand how the part in yellow can give you U_n, I just don't see how taking the two summations away from each other would give U_n, could anyone explain it please
     
  2. jcsd
  3. Jul 24, 2012 #2

    micromass

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    Write it out for n=4. You'll see immediately what happens.
     
  4. Jul 24, 2012 #3

    eumyang

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    Think about it.
    [itex]\Sigma^{n}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n[/itex], and
    [itex]\Sigma^{n - 1}_{r = 1} U_r= U_1 + U_2 + U_3 + ... + U_{n - 1}[/itex].
    So what happens when you subtract the two summations:
    [itex]\left( U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n \right) - \left( U_1 + U_2 + U_3 + ... + U_{n - 1} \right)[/itex]?


    EDIT: Beaten to it. :wink:
     
  5. Jul 24, 2012 #4
    write what out? I've substituted n = 4 and get 20 if I use what they have used for part b...
     
  6. Jul 24, 2012 #5

    eumyang

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    I think micromass meant this:
    n = 4: [itex]\Sigma^{4}_{r = 1} U_r= U_1 + U_2 + U_3 + U_4[/itex]
    n - 1 = 3: [itex]\Sigma^{3}_{r = 1} U_r= U_1 + U_2 + U_3[/itex]
    Don't plug into the expressions with the n's.
     
  7. Jul 24, 2012 #6
    I see, but what is the "n^2 + 4n", is that a general term or..?
     
  8. Jul 24, 2012 #7

    eumyang

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    [itex]\Sigma^{n}_{r = 1} U_r = U_1 + U_2 + U_3 + ... + U_{n - 1} + U_n = n^2 + 4n[/itex]. That was given in the problem. Notice the substitution that was made in the step after the highlighted step.
     
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