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Math Help - Sequences and Series

  1. #1
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    Sequences and Series

    Consider the series:

    1+1/2+1/4+1/8....

    How many terms of this series must be taken before the sum first exceeds 1999/1000?

    What I did was first input 1999/1000 into the series equation, so

    Sn= n/2 (2u1 + (n-1)d)
    1999/1000 = n/2 (2 + (n-1)(1/2) )
    1999/1000 = n (2+ (1/2)n - (1/2) )
    0 = 1/2 n^2 +3/2 n - 1999/500

    Then i used the quad formula and got

    1.7, -4.7

    ... but its wrong lol =|

    i'm not sure what im supposed to do
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  2. #2
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    Quote Originally Posted by juliak View Post
    Consider the series:

    1+1/2+1/4+1/8....

    How many terms of this series must be taken before the sum first exceeds 1999/1000?

    What I did was first input 1999/1000 into the series equation, so

    Sn= n/2 (2u1 + (n-1)d)
    1999/1000 = n/2 (2 + (n-1)(1/2) )
    1999/1000 = n (2+ (1/2)n - (1/2) )
    0 = 1/2 n^2 +3/2 n - 1999/500

    Then i used the quad formula and got

    1.7, -4.7

    ... but its wrong lol =|

    i'm not sure what im supposed to do
    You are using the sum of an arithmetic sequence on a geometric sequence. Try doing the same but using

    S_n = \frac{a(1-r^n)}{1-r}

    Where a is the first term and r is the common ratio
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  3. #3
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    Well, the series is a geometric progression and not arithmetic progression. Hence ur formula for sum is wrong.

    The formula for sum of a geometric progression is:

     Sn = ar^((n-1))/(1-r^n)

    where a is in the initial term in the series
    r is the ratio of the second term to first term or third term to second term and so on....(in this case it is 1/2)
    n is the no. of terms reqd

    The denominator depends on whether r>1 or r<1. In this case, r<1, hence the formula

    Substituting for above parameters and taking the fact that it is just less than 1999/1000, u get,

    (1/2)^((n-1))/(1-(1/2)^n)<1999/1000

    Solve for n.
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  4. #4
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    thank you!

    I simplified down to

    1.9995 < 1/2 ^n

    how can i find the n from that
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  5. #5
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    Quote Originally Posted by juliak View Post
    thank you!

    I simplified down to

    1.9995 < 1/2 ^n

    how can i find the n from that
    ln(1.9995) < -n\, ln(2)

    Note I said that ln(0.5)^n = ln(2^{-1})^n = ln(2^{-n})
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