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Math Help - geometric series...need major help.

  1. #1
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    geometric series...need major help.

    Find the sum of the following infinite
    geometric series. Use a formula and
    show your work.

    (100)+(-40)+(16)+(-32)+etc...forever
    5
    Hint: r = (-2)
    5

    thats -32/5.... and on the hint is -2/5
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  2. #2
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    Do you mean its - 32/5?

    Since the hint says to try r =-2/5, we can plug that in to our values and see what happens
    100(-2/5) = -40
     <br />
-40(-2/5) = 16<br />

    So we see that each term is changing by a factor of -2/5. Now we know the equation for a geometric series is a/(1-r), where a is the first term and r is the ratio. Plug in the first term in the series for a, and plug in the ratio r, and then solve the equation to find the sum of the series.
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  3. #3
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    \sum_{k=0}^{n-1} ar^k=a\frac{1-r^n}{1-r}
    When  \lim_{n\rightarrow \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} since r < 1
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  4. #4
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    so how would i solve that, I dont even know what the E thing is...
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  5. #5
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    Hello, Steph07!

    Find the sum of the following infinite geometric series:
    . . 100 -40 + 16 - \tfrac{32}{5} + \hdots

    You're expected to know the formula . . .

    . . S \;=\;\frac{a}{1-r} . . . where a is the first term and r is the common ratio.


    We have: . a = 100,\;r = \text{-}\tfrac{2}{5}

    Plug them into the formula . . .

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  6. #6
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     <br />
\sum<br />
    means the summation, but I guess you don't really need to know that, just plug into the formula will get you your answer.
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  7. #7
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    so is the answer 100.4 ?
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  8. #8
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    or is it 71.4 ?
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  9. #9
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    I think it should be 71.4.

     <br />
100/(1-(-2/5)) = 100/(1+2/5) = (100/(7/5)) = 500/7<br />
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  10. #10
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    oOo... I see now... thank u
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