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Urgent help required in summation of terms

  1. May 20, 2013 #1
    Find the sum upto n terms:

    1.3.5+3.5.7+5.7.9.............tn

    I solve it this way:

    tn=(2n-1)(2n+1)(2n+3)
    Now can I take summation on both sides? How?

    I mean when I add 2 on both sides the resultant is 0(2-2=0).Similarly the resultant summation will be zero?

    And if I take summation I get one term as 3Ʃ.Now in a book I saw that it is 3n. Why? Summation of 3 will be 3 only as 3 is constant.Please explain.

    I got this:


    Ʃtn=Ʃ(2n-1)(2n+1)(2n+3)

    Ʃtn=Ʃ[(4n^2-1)(2n+3)]
    Ʃtn=Ʃ[8n^3 + 12n^2 - 2n - 3]

    Ʃtn=Ʃ[8n^3] + Ʃ[12n^2] - Ʃ[2n] -Ʃ[3]

    Ʃtn=8*Ʃ[n^3] + 12*Ʃ[n^2] - 2*Ʃ[n] - Ʃ[3]

    Now do I require to write them as Ʃ[3] or 3Ʃ (putting a constant outside Ʃ).Please explain the whole summation process.I am stuck here.
     
  2. jcsd
  3. May 20, 2013 #2
    Hi.

    You are on the right way.

    As for Ʃtn=8*Ʃ[n^3] + 12*Ʃ[n^2] - 2*Ʃ[n] - Ʃ[3], there are formula for Ʃ[n^3], Ʃ[n^2], Ʃ[n] and Ʃ[1].
     
  4. May 20, 2013 #3

    Stephen Tashi

    User Avatar
    Science Advisor

    No. Summing a constant depends on how many times you sum it.

    For example,

    [itex] \sum_{i=1}^4 10 = 10 + 10 + 10 + 10 = (4)(10) = 40 [/itex]
     
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