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Math Help - How do I find the expression for the nth term for these sequences?

  1. #1
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    How do I find the expression for the nth term for these sequences?

    Good Day,

    I'm confused as to how I should derive the expressions for the following sequences. Your explanations and/ or advice is greatly appreciated.

    i) 0.8+0.88+0.888+0.8888+...+(nth term)
    ii) 0.2003+0.20032003+0.200320032003+...+(nth term)

    Regards,
    DD
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  2. #2
    Senior Member Sambit's Avatar
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    (i). 1st term is 8(10)^{-1}

    2nd term is 8(10)^{-1}+8(10)^{-2}

    3rd term is 8(10)^{-1}+8(10)^{-2}+8(10)^{-3}

    and so on.

    So the general expression for nth term will be:

    8(10)^{-1}+8(10)^{-2}+8(10)^{-3}+....+8(10)^{-n}

    = 8[10^{-1}+10^{-2}+....+10^{-n}]. This is a GP series which you can simplify further.


    (ii). 1st term is 2003(10)^{-4}

    2nd term is 2003(10)^{-4}+2003(10)^{-8}

    3rd term is 2003(10)^{-4}+2003(10)^{-8}+2003(10)^{-12}

    and so on.

    So the general expression for nth term will be:

    2003(10)^{-4}+2003(10)^{-8}+2003(10)^{-12}+....+2003(10)^{-4n}

    = 2003[10^{-4}+10^{-8}+....+10^{-4n}]
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  3. #3
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    Hello, dd86!

    (1)\;\;0.8+0.88+0.888+0.8888+ \hdots + (n^{th}\text{ term})

    We have: . 0.8 + 0.88 + 0.888 + 0.8888 + \hdots

    . . . . . =\;0.8(1 + 1.1 + 1.11 + 1.111 + \hdots)


    \text{The }n^{th}\text{ term is: }\:a_n \;=\;(0.8)(\underbrace{1.111\hdots 1}_{\text{geometric series}})

    . . 1.111\hdots1 \:=\:\dfrac{1-(0.1)^n}{1-0.1} \;=\; \dfrac{1-(0.1)^n}{0.9}


    \text{Therefore: }\;a_n \;=\;(0.8)\,\left[\dfrac{1-(0.1)^n}{0.9}\right] \;=\;\frac{8}{9}\,\left[1 - (0.1)^n\right]




    (2)\;\;0.2003+0.2003\,2003+0.2003\,2003\,2003+ \hdots + (n^{th}\text{  term})

    We have: . 0.2003\,\left(1 + 1.0001 + 1.00010001 + 1.000100010001 + \hdots\right)

    \text{The }n^{th}\text{ term is: }\;a_n \;=\;0.2003\,(1.0001\,0001,\,0001 \hdots 0001)

    . . \displaystyle 1.00010001\hdots 0001 \;=\;\frac{1-(0.0001)^n}{1- 0.0001} \;=\;\frac{1-(0.0001)^n}{0.9999}


    \text{Therefore: }\;a_n \;=\;(0.2003)\,\dfrac{1-(0.0001)^n}{0.9999} \;=\;\frac{2003}{9999}\left[1 - (0.0001)^n\right]

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  4. #4
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    Thank you all for your replies. It's clearer to me now.

    However, what should I do to get obtain the summation of the series up to the nth term?

    I notice a pattern when I sum the first few terms up. The issue is that I can't find a way to express it in a general manner.
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  5. #5
    Senior Member Sambit's Avatar
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    This is a Geometric Series. a+ar+ar^2+ar^3+....+ar^{n-1} can be simplified as a\frac{r^n-1}{r-1}
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