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Math Help - Sum of the sum of an arithmetic progression

  1. #1
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    Sum of the sum of an arithmetic progression

    a sequence is given as: 3,9,18,30,45....... (i will call this sequence A)
    for this sequence the sum of the first 100 terms is needed

    here is what i have done:
    the difference of the above sequence is an arithmetic progression
    ie: 3,6,9,12,15,........... (i will call this sequence B)

    using some formula manipulation i have come up with an equation for the sequence 3,9,18,30,45,...................

    which is
    (n/4)*[(n+1)*(2a+(n-1)*d)+ (d/3) - ((n^2*d)/3)]

    where n is the number of terms needed to be added

    a is the first term of the sequence B (or sequence A since both first terms are the same)

    d is the common difference in sequence B

    using this formula i can come up with the answer but the formula is rather lengthy.....is there anyone who has a simpler formula or someone who can further simplify this one.
    Thanks in advance
    swnz
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  2. #2
    MHF Contributor red_dog's Avatar
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    a_1=3=3\cdot 1

    a_2=9=3(1+2)

    a_3=18=3(1+2+3)

    a_4=30=3(1+2+3+4)

    a_5=45=3(1+2+3+4+5)

    Then a_n=3(1+2+3+\ldots+n)=3\cdot\frac{n(n+1)}{2}=\frac  {3}{2}(n^2+n)

    \sum_{k=1}^{100}a_k=\frac{3}{2}\sum_{k=1}^{100}(k^  2+k)=\frac{3}{2}\left(\sum_{k=1}^{100}k^2+\sum_{k=  1}^{100}k\right)

    Now use the formula \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6} and \sum_{k=1}^nk=\frac{n(n+1)}{2} replacing n with 100.
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