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Sigma Notation problem

  1. Oct 19, 2007 #1
    [a]1. The problem statement, all variables and given/known data[/b]

    [tex]\sum[/tex] (6n + 1)

    2. The attempt at the solution
    So, how do i go about doing this? I tried finding the first three numbers of the series to find the difference by substituting n = 5,6,7 and then use the Sum formula S = n/2[2a + (n-1)d]. But the answer turned out to be wrong. The correct answer to this is 416.
    Last edited: Oct 19, 2007
  2. jcsd
  3. Oct 19, 2007 #2
    Remember that you are summing from n=5 to n=12, not n=0 to n=12. Does that make a difference? Can't really tell because tyou never showed your working.
  4. Oct 19, 2007 #3
    Sorry about that.
    So i substituted n = 5,6,7
    so the sequences is as follows: 31,37,43.
    So from the sequence a = 31 and d = 6
    Substituting into the formula S = n/2[2a + (n-1)d]

    S = 12/2 [2(31) + (12-1)6]
    S = 6[62 + 66]
    S = 6 X 128
    S = 768
    The answer i get is pretty farout from the actual answer.

    EDIT : Found the solution. the number of terms is 8, not 12. Because you actually count the number of terms from 5 to 12.
    Last edited: Oct 19, 2007
  5. Oct 19, 2007 #4
    Correct, you should remember that if you are summing from a to b there are (b-a+1) terms. If you take (b-a) terms you will be one short because you have not included a as your first term. Examine this, think of the difference as being equal to the number of bracketed things (if that makes sense):

    b-a = (a+1),(a+2),(a+3),...(b)
    b-a+1= [(a+1),(a+2),(a+3),...(b)]+1
    b-a+1= (a),(a+1),(a+2),(a+3),...,(b)

    Do you see why it should be b-a+1 and not b-a ?

    I am sorry if I am labouring the point too much :(
  6. Oct 19, 2007 #5
    Yes, got the point. Thanks a lot :D
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