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Sum of r(r+1)

  1. Nov 27, 2012 #1
    Calculate 5 [itex]\Sigma[/itex] r=0 r(r+1)
    (Sorry I don't know how to do the proper notation online)

    How do you calculate the sum for this since the common difference is changing?
    I tried to write them out separately so the sum of r x sum of r+1 but I don't know how you put that in the sum formula.

    Sn = n/2[2a + (n-1)d]
    Sn = n/2[a + L]
     
  2. jcsd
  3. Nov 27, 2012 #2

    haruspex

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    Hi Nubcake,
    Given terms that grow like r2, how fast do you think the sum will grow?
     
  4. Nov 27, 2012 #3

    Mark44

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    I don't know what you're asking.
     
  5. Nov 27, 2012 #4

    SammyS

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    Do you mean [itex]\displaystyle \sum_{r=0}^{5}r(r+1)\ ?[/itex]
     
  6. Nov 27, 2012 #5

    Curious3141

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    Assuming you meant the sum as SammyS wrote it, there are at least a couple of ways to do it.

    The most elementary way, but one that involves making a non-trivial observation, is to use what's called the "method of differences". Observe that [itex](r+1)^3 - r^3 = 3r(r+1) + 1[/itex] What happens if you write out the sum of the left hand side? What terms cancel?

    The less elementary way, but one that's easier to see if you know just a little calculus, is to let [itex]f(x) = x^{r+1}[/itex]. Now differentiate twice, and work out the value at x = 1, i.e. calculate [itex]f''(1)[/itex]. What do you notice?
     
  7. Nov 27, 2012 #6

    Mark44

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    Given that this was posted in the Precalc section, a solution involving calculus might not be applicable here.
     
  8. Nov 27, 2012 #7

    HallsofIvy

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    The most obvious way to do [itex]\sum_{r=0}^5 r(r+1)[/itex] is to write out all six terms and sum them: 0(0+1)+ 1(1+1)+ 2(2+1)+ 3(3+1)+ 4(4+1)+ 5(5+ 1)= 0+ 2+ 6+ 12+ 20+ 30= 70.
     
  9. Nov 27, 2012 #8

    Curious3141

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    That does not mean the OP has no knowledge of it. Besides, it's better to suggest alternative solutions, especially if they're more direct and easier to see. The OP is free to disregard this solution if he so wishes.
     
  10. Nov 28, 2012 #9

    Mark44

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    I realize that problems are posted to the wrong sections all the time, and that is why I qualified what I said. On the other hand, the OP did post this in the precalc section, so I take that as a clue to the OP's current abilities. A calculus approach might seem more direct and easier to those of us with more knowledge, but would probably be completely mystifying to someone who isn't ready for such an approach yet.
     
  11. Nov 28, 2012 #10

    Curious3141

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    I think it's pointless to keep going on about this. It's really for the OP to reply, choose a method and show more work.
     
  12. Nov 28, 2012 #11

    Mark44

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    I agree completely.
     
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