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Math Help - Is it true that...

  1. #16
    Super Member Bacterius's Avatar
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    If what you wrote is what I read, then yes, I am absolutely sure.
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  2. #17
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    Quote Originally Posted by wonderboy1953 View Post
    Sorry I made the problem too hard:

    1/(\sqrt{1} + \sqrt{2}) + 1/(\sqrt{2} + \sqrt{3}) + 1/(\sqrt{3} + \sqrt{4})+...+1/(\sqrt{498} + \sqrt{499}) + 1/(\sqrt{499} + \sqrt{500}) = 499?

    It's a great problem, algebra wise (that's a hint) and how would you go about proving that the left side of the equation equals the right side?
    <br />
\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =<br />

    <br />
 = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =<br />

    <br />
 = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1<br />

    Maybe this is what you meant?
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  3. #18
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    Quote Originally Posted by Unbeatable0 View Post
    <br />
\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =<br />

    <br />
 = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =<br />

    <br />
 = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1<br />

    Maybe this is what you meant?
    Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).
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  4. #19
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    Quote Originally Posted by Unbeatable0 View Post
    <br />
\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =<br />

    <br />
 = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} = \sum_{k=1}^n \frac{1}{\sqrt{k+1} + \sqrt{k}} \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1} - \sqrt{k}} =<br />

    <br />
 = \sum_{k=1}^n (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}-1<br />

    Maybe this is what you meant?
    Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).
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  5. #20
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    Quote Originally Posted by wonderboy1953 View Post
    Are you sure about this, Wilmer and Bacterius? Please recheck
    YES.

    Methinks YOU should recheck.
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  6. #21
    Super Member Deadstar's Avatar
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    Quote Originally Posted by wonderboy1953 View Post
    Sorry I made the problem too hard:

    1/(\sqrt{1} + \sqrt{2}) + 1/(\sqrt{2} + \sqrt{3}) + 1/(\sqrt{3} + \sqrt{4})+...+1/(\sqrt{498} + \sqrt{499}) + 1/(\sqrt{499} + \sqrt{500}) = 499?

    It's a great problem, algebra wise (that's a hint) and how would you go about proving that the left side of the equation equals the right side?
    LHS = 21.36067983

    It's pretty clear that it's not 499.

    Every term in the sum is less than 1 and there are 499 of them.
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  7. #22
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    Quote Originally Posted by Deadstar View Post
    LHS = 21.36067983

    It's pretty clear that it's not 499.

    Every term in the sum is less than 1 and there are 499 of them.
    Deadstar, I had set up the problem wrong (but it's a great problem).
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  8. #23
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    Well.....what d'heck is the problem when set up RIGHT?
    Why don't you post it again CORRECTLY, on a NEW thread, so we don't get all mixed up here!!
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  9. #24
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    Quote Originally Posted by Wilmer View Post
    Well.....what d'heck is the problem when set up RIGHT?
    Why don't you post it again CORRECTLY, on a NEW thread, so we don't get all mixed up here!!
    Two reasons: I don't want to repeat and I'm tired (at my library I have a time constraint which helps to wear me out).
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  10. #25
    Super Member Bacterius's Avatar
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    Two reasons: I don't want to repeat and I'm tired (at my library I have a time constraint which helps to wear me out).
    Repeat what ? You haven't even given the "great" problem yet.
    Never mind, just post when you feel like it. And get it right
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  11. #26
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    Quote Originally Posted by wonderboy1953 View Post
    Yes, that's it Unbeatable0. Part of the point to this problem is that it's really not as hard as it looks and rationalizing the denominator leads to cancellation of terms and the solution (I should have picked n = 528).
    Dear Wonderboy1953,

    Well I don't understand. If you want \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{n+1}-1=499

    Then \sqrt{n+1}=500\Rightarrow{n=249999}

    If you say that you should have picked n=528 then,

    \frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + ... + \frac{1}{\sqrt{n-1} + \sqrt{n}} + \frac{1}{\sqrt{n} + \sqrt{n+1}} =\sqrt{528+1}-1=22

    Please clarify!!
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  12. #27
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    Quote Originally Posted by Bacterius View Post
    Repeat what ? You haven't even given the "great" problem yet.
    Never mind, just post when you feel like it. And get it right
    (Forgive me moderators as I'm commanded to repeat and to get it right)

    Is it true that:

    1/(\sqrt1 + \sqrt2) + 1/(\sqrt2 + \sqrt3) + 1/(\sqrt3 + 1\sqrt4)+...+1/(\sqrt483 + \sqrt484) = 21?
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  13. #28
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    Quote Originally Posted by wonderboy1953 View Post
    (Forgive me moderators as I'm commanded to repeat and to get it right)
    Is it true that:
    1/(\sqrt1 + \sqrt2) + 1/(\sqrt2 + \sqrt3) + 1/(\sqrt3 + 1\sqrt4)+...+1/(\sqrt483 + \sqrt484) = 21?
    May I ask, Wboy, why does "posting it right" seem unimportant to you?

    Anyway, answer to your "CORRECT" posting above is YES!!
    Would be better if this way:
    1 / [sqrt(0) + sqrt(1)] + 1 / [sqrt(1) + sqrt(2)] + ..... + 1 / [sqrt(483) + sqrt(484)] = 22

    Get my drift? Answer is the square root of last number (484 in this case)

    In other words:
    1 / [sqrt(0) + sqrt(1)] + 1 / [sqrt(1) + sqrt(2)] + ..... + 1 / [sqrt(n-1) + sqrt(n)] = sqrt(n)

    Now I kinda understand why it's GREAT (when correct, of course!)
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