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Babynini

  • one year ago

A partial sum of an arithmetic sequence is given. find the sum. 1+5+9+...+401

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  1. anonymous
    • one year ago
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    ok, we have \(a_1=1\) and common difference is \(d=4\), last term is \(a_n=401\) and you sure know that nth term of the sequence (a_n) is given by:\[a_n=a_1+(n-1)d\]find \(n\)\[401=1+4(n-1) \\ n-1=100 \\ n=101\]and for sum use the formula\[S=\frac{n(a_1+a_n)}{2}=\frac{101(1+401)}{2}=101\times 201\]

  2. Babynini
    • one year ago
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    What if it is a geometric sum?

  3. Babynini
    • one year ago
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    For example: 1+3+9+...+2187 (thanks for the help on the other one by the way :) )

  4. Babynini
    • one year ago
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    @mukushla :)

  5. Babynini
    • one year ago
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    from that we get, a=, r=3, n = ?

  6. Babynini
    • one year ago
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    Plugging some numbers here and there I get that a_8=2187 now what?

  7. Babynini
    • one year ago
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    nvm, I got it now xD it = 3280

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