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anonymous

  • one year ago

A = (1+2)/3 + (1+2+3)/3^2 + (1+2+3+4)/3^3 + ... The value of 8A = ....

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  1. anonymous
    • one year ago
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    The choices : A. 3 B. 15 C. 19 D. 22

  2. ganeshie8
    • one year ago
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    Recall the geometric series : \[\sum\limits_{n=0}^{\infty}x^n = \dfrac{1}{1-x}\] differentiate and get \[\sum\limits_{n=1}^{\infty}nx^{n-1} = \dfrac{-1}{(1-x)^2}\] differentiate again and get \[\sum\limits_{n=2}^{\infty}n(n-1)x^{n-2} = \dfrac{2}{(1-x)^3}\] plugin \(x=\dfrac{1}{3}\) and massage a bit

  3. imqwerty
    • one year ago
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    the whole sum A can be represented as - \[\sum_{n=1}^{\infty}\frac{ (n+1)^2 }{ 2(3^n) }\] can u evaluate this thing by breaking into parts :)

  4. anonymous
    • one year ago
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    @imqwerty unless I'm missing something, did you mean \[\sum_{n\ge1}\frac{n(n+1)}{2(3^n)}~~?\]

  5. imqwerty
    • one year ago
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    the numerator has n+1 terms with the 1st term=1 and common diff=1 so the numerator wuld be (n+1)(2+(n-1))/2 =(n+1)(n+1)/2=(n+1)^2/2

  6. anonymous
    • one year ago
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    Hmm, the reason I'm confused is that the numerators belong to the sequence of sums of consecutive positive integers, i.e. \[\sum_{i=1}^ki=1+2+\cdots+k,\quad \text{for }k\ge2\] This sum has a nice closed form \[\sum_{i=1}^ki=\binom k2=\frac{k(k+1)}{2}\] so the original series should be \[\frac{1+2}{3}+\frac{1+2+3}{3^2}+\cdots=\sum_{n=1}^\infty \frac{\displaystyle\sum_{i=1}^{n+1}i}{3^n}=\sum_{n=1}^\infty \frac{(n+1)(n+2)}{2(3^n)}\]

  7. anonymous
    • one year ago
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    (Minor typo in my first comment)

  8. anonymous
    • one year ago
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    @ganeshie8, how to massage it ? ._.

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