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Math Help - a series problems

  1. #1
    Member zzizi's Avatar
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    a series problems

    Hi would appreciate some help in how i can solve this problem..




    i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
    to become greater than 80.



    thanks in advance
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  2. #2
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    Quote Originally Posted by zzizi View Post
    Hi would appreciate some help in how i can solve this problem..




    i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
    to become greater than 80.



    thanks in advance
    Go to the formula for the sum of a geometric series. Substitute that the first term is 0.121 and the common ratio is 2. Put it greater than 80 and solve the resulting inequality for the minimum value of n.
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  3. #3
    Member zzizi's Avatar
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    MrFantastic thank you for the reply.

    Pls correct me if this is wrong...

    should i do this?

    S3 = 0.121(1-2^3)/ (1-2)>80
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  4. #4
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    Quote Originally Posted by zzizi View Post
    MrFantastic thank you for the reply.

    Pls correct me if this is wrong...

    should i do this?

    S3 = 0.121(1-2^3)/ (1-2)>80
    Please post the formula for the sum of a geometric series and then show how you used it to get the above.
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  5. #5
    Member zzizi's Avatar
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    ok... this is the formula i just used
    S(subN) =a(1-r^n)/1-r

    and then i set it greater than 80

    should i use this formula or is there another one?
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  6. #6
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    Quote Originally Posted by zzizi View Post
    ok... this is the formula i just used
    S(subN) =a(1-r^n)/1-r

    and then i set it greater than 80

    should i use this formula or is there another one?
    Yes. So 0.121(1-2^3)/ (1-2)>80 (post #3) should be 0.121(1-2^n)/ (1-2)>80 and you solve for the minimum integer value of n.
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  7. #7
    Member zzizi's Avatar
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    Oh right thanks.... so then the inequality would require me to use logs to solve? ... i will post my working out in a min...
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  8. #8
    Member zzizi's Avatar
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    i cant type up all the symbols... but what i tried to do was the following:
    0.121-2^n/1-2 >80
    then i subtracted 2 from the denominator which left -1 .. i then divided both sides by neg 1 where switched signs to this
    0.121-2^n<-80
    then added 0.121 to both sides
    -2^n<-79.9
    i then to log of both side
    n(ln)-2<(ln)-79.9
    to get this
    n< (ln) -79.9/ (ln) -2
    therefore
    n<6.3
    pls tell me this right?


    anyone?
    Last edited by zzizi; April 20th 2010 at 06:45 AM.
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  9. #9
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    Quote Originally Posted by zzizi View Post
    i cant type up all the symbols... but what i tried to do was the following:
    0.121-2^n/1-2 >80
    then i subtracted 2 from the denominator which left -1 .. i then divided both sides by neg 1 where switched signs to this
    0.121-2^n<-80
    then added 0.121 to both sides
    -2^n<-79.9
    i then to log of both side
    n(ln)-2<(ln)-79.9
    to get this
    n< (ln) -79.9/ (ln) -2
    therefore
    n<6.3
    pls tell me this right?


    anyone?
    Substitute your answer into the inequality. Does it work? Note: n is meant to be an integer. And IF your answer is correct, it should read n > 6.3 ....

    Personally, I'd solve the inequality using trial-and-error.
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  10. #10
    Member zzizi's Avatar
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    yes i think it does work.. because i tried adding up 6 terms and it does sum to over 80... i need to brush up on this area.. thanks very much for help.
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  11. #11
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    Quote Originally Posted by zzizi View Post
    yes i think it does work.. because i tried adding up 6 terms and it does sum to over 80
    [snip]
    Really? Please show the details ....
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  12. #12
    Member zzizi's Avatar
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    oh.... erm... ok now im alittle confused... what i meant was that i multiplied 0.121 six times by 2 in the calculator and it added up to over 80.
    pls do shed some light as to how i should find the answer and was my working out the right way to solve this problem?
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  13. #13
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    Quote Originally Posted by zzizi View Post
    oh.... erm... ok now im alittle confused... what i meant was that i multiplied 0.121 six times by 2 in the calculator and it added up to over 80.
    [snip]
    Since 2^6 = 64 I doubt very much that 0.121 multiplied six times by 2 will be greater than 80.

    I have told you how to do the question, but until you can do the basic arithmetic correctly (and I'm not going to that for you) you will not get the correct answer.

    (It is simple enough to write out and then add up the first six terms - doing so should have quickly shown you that your answer was wrong).
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  14. #14
    Member zzizi's Avatar
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    ok point taken... thanks
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  15. #15
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    Quote Originally Posted by zzizi View Post
    Hi would appreciate some help in how i can solve this problem..




    i have to find the minimum number of terms which would make the sum of the series 0.121+0.242+0.484 ...
    to become greater than 80.



    thanks in advance
    The formula you used above is good, and in case you're interested here's another way of looking at it:

    0.121+0.242+0.484+...

    =0.121(1+2+4+8+...)

    Consider the series

    1+2+4+8+...

    and the sequence of its partial sums

    \{a_n\}=1,3,7,15,31,...

    Notice that a_n=2^n-1

    So we reach the same conclusion that 0.121(2^n-1)>80

    Like mr fantastic mentioned, trial and error is pretty easy here, especially if you are familiar with powers of 2 since they come up a lot.

    But if you feel you absolutely must solve the inequality exactly, then I can show you the first couple steps

    0.121\left(2^n-1\right)>80

    2^n-1>\frac{80}{0.121}

    2^n>\frac{80}{0.121}+1

    You will end up using logarithms.. well I practically did the problem for you already. Anyway good luck. By the way there is a forum that discusses math formatting using LaTeX.
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