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Math Help - can anybody help me solve the value of this...

  1. #1
    rcs
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    can anybody help me solve the value of this...

    Please help me on this ... thanks
    Attached Thumbnails Attached Thumbnails can anybody help me solve the value of this...-mathxxx.jpg  
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  2. #2
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    Have you tried actually evaluating the stuff under the square roots? The first three are perfect squares... I suspect there's a pattern...
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    rcs
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    i solved for the first 3 but will we know it? they only gave the last term after the first 3 terms... too confusing
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    The general term appears to be \displaystyle \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}

    \displaystyle = \sqrt{\frac{n^2(n+1)^2}{n^2(n + 1)^2} + \frac{(n + 1)^2}{n^2(n+1)^2} + \frac{n^2}{n^2(n + 1)^2}}

    \displaystyle = \sqrt{\frac{n^2(n+1)^2 + (n + 1)^2 + n^2}{n^2(n + 1)^2}}

    \displaystyle = \sqrt{\frac{n^2(n^2 + 2n + 1) + n^2 + 2n + 1 + n^2}{n^2(n+1)^2}}

    \displaystyle = \sqrt{\frac{n^4 + 2n^3 + n^2 + 2n^2 + 2n + 1}{n^2(n+1)^2}}

    \displaystyle = \sqrt{\frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2}}

    \displaystyle = \sqrt{\frac{(n^2 + n +1)^2}{n^2(n+1)^2}}

    \displaystyle = \frac{n^2 + n + 1}{n(n + 1)}

    \displaystyle = \frac{n^2 + n + 1}{n^2 + n}

    \displaystyle = 1 + \frac{1}{n^2 + n}

    \displaystyle = 1 + \frac{1}{n(n + 1)}

    \displaystyle = 1 + \frac{1}{n} - \frac{1}{n + 1}.



    So your series is

    \displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right).

    This is a telescopic series... See how you go evaluating it...
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  5. #5
    rcs
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    too long to be manipulated by summation... thanks anyway
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    Quote Originally Posted by rcs View Post
    too long to be manipulated by summation... thanks anyway
    Oh no it is not. It is so simple.

    \displaystyle \sum_{n = 1}^{1999}\left(1 + \frac{1}{n} - \frac{1}{n + 1}\right)= \displaystyle\sum_{n = 1}^{1999}1+\sum_{n = 1}^{1999}\left(\frac{1}{n} - \frac{1}{n + 1} \right)= 1999+1-\frac{1}{2000}
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    Quote Originally Posted by rcs View Post
    too long to be manipulated by summation... thanks anyway
    I see - well next time you post a difficult looking problem I'll just say it'll take up too much of my time, considering you're not even going to put in any effort...
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  8. #8
    rcs
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    i would like to apologize to what i had replied on Prove it's answer... i never meant to be so mean... my point was, i didn't understand what he had been solving, i mean the answer itself is too difficult for me to understand , i was thinking that i would be solving too long like adding from 1 to 1999 consecutive numbers because it is summation ... i didn't know how it would go to 1999 + 1 - 1/2000 .... sorry i really cannot get how it would become like that... sorry hope you guys understand... it was just miscommunication.

    thanks
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    Do you know what "telescoping series" means?

    Prove It showed that the series can be written
    \displaytype \sum_{n=1}^{1999}\left(1+ \frac{1}{n}- \frac{1}{n+1}\right)

    That is,
    (1+ 1- \frac{1}{2})+ (1+ \frac{1}{2}- \frac{1}{3})+ (1+ \frac{1}{3}+ \frac{1}{4}+ (1+ \frac{1}{4}- \frac{1}{5}) \cdot\cdot\cdot

    Do you see how the fractions cancel? All except in the last term, -\frac{1}{2000}. Of course, you will also have "1" added 1999 times.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    Do you know what "telescoping series" means?

    Prove It showed that the series can be written
    \displaytype \sum_{n=1}^{1999}\left(1+ \frac{1}{n}- \frac{1}{n+1}\right)

    That is,
    (1+ 1- \frac{1}{2})+ (1+ \frac{1}{2}- \frac{1}{3})+ (1+ \frac{1}{3}+ \frac{1}{4}+ (1+ \frac{1}{4}- \frac{1}{5}) \cdot\cdot\cdot

    Do you see how the fractions cancel? All except in the last term, -\frac{1}{2000}. Of course, you will also have "1" added 1999 times.
    I'm sure you mean

    \displaystyle \left(1 + 1 - \frac{1}{2}\right) + \left(1 + \frac{1}{2} - \frac{1}{3}\right) + \left(1 + \frac{1}{3} - \frac{1}{4}\right) + \left(1 + \frac{1}{4} - \frac{1}{5}\right) + \dots + \left(1 + \frac{1}{1999} - \frac{1}{2000}\right).

    But yes, a telescopic series means that all terms except for the first and last end up cancelling.
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    Yes, thanks. I accidently changed a "-" to a "+".
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  12. #12
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    thanks

    honestly i really dont know what telescoping series mean..
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  13. #13
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    Quote Originally Posted by rcs View Post
    honestly i really dont know what telescoping series mean..
    Many textbooks call them collapsing sums.
    Look at \left[ {1 - \frac{1}{2}} \right] + \left[ {\frac{1}{2} - \frac{1}{3}} \right] + \left[ {\frac{1}{3} - \frac{1}{4}} \right] + \left[ {\frac{1}{4} - \frac{1}{5}} \right] + \left[ {\frac{1}{5} - \frac{1}<br />
{6}} \right].
    Remove the brackets.
    Every term subtracts off, collapses, except the first and last.
    So we have  \sum\limits_{k = 1}^5 {\left[ {\frac{1}{k} - \frac{1}{{k + 1}}} \right]}  = 1 - \frac{1}{6}
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  14. #14
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    Thanks Everybody!
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