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Math Help - geometric sequences - closed forms

  1. #1
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    geometric sequences - closed forms

    i have just started my OU course and am struggling to get to terms with the geometric sequences, i believe i cannot post the actual question on here, but was wondering if anyone can help me with explaining how i can find a closed form, i have a regular pattern of deducting 0.7 from each previous answer, so i know it is an arithmetic sequence, but just cannot get to grips with the closed form

    cheers guys
    trace
    Last edited by tracey0245; February 13th 2010 at 05:42 AM. Reason: titlew wrong
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  2. #2
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    Quote Originally Posted by tracey0245 View Post
    i believe i cannot post the actual question on here,
    Why is that the case?
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  3. #3
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    Quote Originally Posted by tracey0245 View Post
    i have just started my OU course and am struggling to get to terms with the geometric sequences, i believe i cannot post the actual question on here, but was wondering if anyone can help me with explaining how i can find a closed form, i have a regular pattern of deducting 0.7 from each previous answer, so i know it is an arithmetic sequence, but just cannot get to grips with the closed form

    cheers guys
    trace
    I guess I am a little confused about your question. First you state that it is a geometric series. This means you are multiplying by the same ratio every time an example would be

    4+2+1+\frac{1}{2}+\frac{1}{4}+...=\sum_{n=0}^{\inf  ty}4\left( \frac{1}{2}\right)^{n}

    Where the common ratio r=\frac{1}{2} and your starting value is 4.

    So I guess my question for you is when you say
    i have a regular pattern of deducting 0.7 from each previous answer
    By decucting do you mean subtraction or are you multiplying by 0.7
    If you are multiplying the geometric series should look like

    \sum_{n=0}^{\infty}ar^{n}=\frac{a}{1-r}

    Or is the sum is finite you would get

    \sum_{n=0}^{k-1}ar^{n}=\frac{a(1-r^k)}{1-r}

    P.S what is OU?
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    P.S what is OU?
    a second tier university in Norman, Oklahoma ... with "land thieves" for a mascot.

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  5. #5
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    still trying lol

    i said geometric series, as that what is comes under i think? the open university(OU) states that i cannot display my questions,and i don't want to cheat at all, hence why i asked for help with how to figure it out
    i have done a lot of reading it is an arithmetic sequence, that starts from 3.2 and follows decreases by 0.7 each time, (starts at 3.2)
    i have found the recurrent system to descrtibe it, however i am struggling to find the closed form 4 this.
    I have been pickling it for ages and the closest i get is :
    3.2=(Un-1)0.7
    or Un=3.2-(0.7)n-1


    hope this is a bit clearer, cheers
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  6. #6
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    Quote Originally Posted by tracey0245 View Post
    i said geometric series, as that what is comes under i think? the open university(OU) states that i cannot display my questions,and i don't want to cheat at all, hence why i asked for help with how to figure it out
    i have done a lot of reading it is an arithmetic sequence, that starts from 3.2 and follows decreases by 0.7 each time, (starts at 3.2)
    i have found the recurrent system to descrtibe it, however i am struggling to find the closed form 4 this.
    I have been pickling it for ages and the closest i get is :
    3.2=(Un-1)0.7
    or Un=3.2-(0.7)n-1


    hope this is a bit clearer, cheers
    Actually, that makes it less clear! You now talk about "arithmetic sequence" as well as "geometric series and also talk about "recurrence" when neither of those last equations is a recurrence relation! The first does not have a " U_n and the second does not have a U_{n-1}.

    For a geometric sequence, a r^n, where each term is the previous term multiplied by a constant, your recurrence relation would be img.top {vertical-align:15%;} U_n= 0.7U_n with U_0= 3.2 and the closed form U_n= (3.2)(0.7)^n

    For an arithmetic sequence, a+ id, where each term is the previous term with a constant, d, added, your recurence relation would be U_n= U_{n-1}+ d. Here that would be [maht]U_n= U_{n-1}- 0.7" alt="U_n= rU_{n-1}. Here that would be U_n= 0.7U_n with U_0= 3.2 and the closed form U_n= (3.2)(0.7)^n

    For an arithmetic sequence, a+ id, where each term is the previous term with a constant, d, added, your recurence relation would be U_n= U_{n-1}+ d. Here that would be [maht]U_n= U_{n-1}- 0.7" /> and U_0= 3.2 and the closed form U_n= 3.2- 0.7 n.

    A "series" would be a sum of those things.
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  7. #7
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    arithmetic sequence

    cheers for the help hallsofivy, yeah sorry for the confusion, it was an arithmetic sequence. What i got confused on is my next one is a geometric sequence, which i need to find the first four terms and a closed sequence for, if i show you the sequence can you not tell me the answer please but just some guidance on how i should be working it out, cos never seen this sort of geometric sequence before.

    u1=5, un+1=0.4un (n=1,2,3....)

    many thanks
    trace
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