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Math Help - Series Problem

  1. #1
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    Series Problem

    Find:
    1 + 2a + 3a + 4a + ... + na^(n-1)
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  2. #2
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    Hello, bearej50!

    Find: . 1 + 2a + 3a^2  + 4a^3 + \hdots + na^{n-1}

    . . . . \text{We have: }\quad S \;=\;  1+2a + 3a^2 + 4a^3 + \hdots + na^{n-1}

    \text{Multiply by }a\!:\;\;\;aS \;=\;\qquad a + 2a^2 + 3a^3 + \hdots + (n-1)a^n + na^{n+1}


    \text{Subtract: }\;S - aS \;=\;\underbrace{1 + a + a^2+ a^3 + a^n}_{\text{geometric series}} - na^{n+1}

    . . . (1-a)S \;=\;\frac{1-a^n}{1-a} - na^{n+1} \;=\;\frac{1-a^n - na^{n+1}(1-a)}{1-a}


    \text{Therefore: }\;S \;=\;\frac{1 - a^n - na^{n+1}(1-a)}{(1-a)^2}

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  3. #3
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    Gothenburg, Sweden
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    Let f(a)=1+2a + 3a^2 + 4a^3 + \hdots + na^{n-1}. Integrate both sides, this is F(a)=a+a^2+ \hdots a^n=\frac{1-a^{n+1}}{1-a}-1. Differentiating yields the result
    f(a)=\frac{1-a^{n+1}}{(1-a)^2}-\frac{(n+1)a^n}{1-a}=\frac{1-a^n-na^n+na^{n+1}}{(1-a)^2} and this is your answer.
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