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Math Help - Square Root of Sum of the Numbers

  1. #1
    Member Rimas's Avatar
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    Square Root of Sum of the Numbers

    What is the square root of the sum of the first 2007 positive odd integers?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Rimas View Post
    What is the square root of the sum of the first 2007 positive odd integers?

    \sqrt{\sum^{2007}_{n=1}(2n-1)}

    =\left(2\sum^{2007}_{n=1}(n)-\sum^{2007}_{n=1}(1)\right)^\frac{1}{2}

    =\left(2\left(\frac{2007(2007+1)}{2}\right)-(2007)\right)^\frac{1}{2}

    =\left(2007(2007+1-1)\right)^\frac{1}{2}

    =\left((2007)^2\right)^\frac{1}{2}

    =2007
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  3. #3
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    Hello, Rimas!

    This could be a trick question . . .


    What is the square root of the sum of the first 2007 positive odd integers?
    We have: . S \;= \;\underbrace{1 + 3 + 5 + 7 + \cdots }_{2007\text{ terms}}

    This is an arithmetic series with first term a = 1 and common difference d = 2

    The sum of the first n terms is: . \frac{n}{2}[2a + (n-1)d]

    So we have: . S \;=\;\frac{2007}{2}[2\cdot1 + 2006(2)] \;=\;2007^2

    Therefore: . \sqrt{S}\;=\;\sqrt{2007^2} \;=\;2007

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If you knew the following bit of mathematical trivia,
    . . you could have answered it immediately.

    \begin{array}{ccccc}1 & = & 1 & = & 1^2 \\1 + 3 & = & 4 & = & 2^2 \\1 + 3 + 5 & = & 9 & = & 3^2 \\1 + 3 + 5 + 7 & = & 16 & = & 4^2 \\1 + 3 + 5 + 7 + 9 & = & 25 & = & 5^2\end{array}

    Get it?
    The sum of the first n odd number is always n^2.

    So the sum of the first 2007 odd number is 2007^2.

    And its square root is, of course, 2007.

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Rimas!

    This could be a trick question . . .


    We have: . S \;= \;\underbrace{1 + 3 + 5 + 7 + \cdots }_{2007\text{ terms}}

    This is an arithmetic series with first term a = 1 and common difference d = 2

    The sum of the first n terms is: . \frac{n}{2}[2a + (n-1)d]

    So we have: . S \;=\;\frac{2007}{2}[2\cdot1 + 2006(2)] \;=\;2007^2

    Therefore: . \sqrt{S}\;=\;\sqrt{2007^2} \;=\;2007

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If you knew the following bit of mathematical trivia,
    . . you could have answered it immediately.

    \begin{array}{ccccc}1 & = & 1 & = & 1^2 \\1 + 3 & = & 4 & = & 2^2 \\1 + 3 + 5 & = & 9 & = & 3^2 \\1 + 3 + 5 + 7 & = & 16 & = & 4^2 \\1 + 3 + 5 + 7 + 9 & = & 25 & = & 5^2\end{array}

    Get it?
    The sum of the first n odd number is always n^2.

    So the sum of the first 2007 odd number is 2007^2.

    And its square root is, of course, 2007.

    Thanks Soroban. I was thinking of doing it ecMathGeeks way, but never considered arithmetic series. That trivia was nice as well
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  5. #5
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    Hello, ecMathGeek!

    A straight-forward, no-nonsense approach . . . great job!

    I intended to use your method, but something bothered me.

    The square root of the sum? . . . Won't that be some ugly decimal?
    . . Why in the world would they want such a . . . Oh!

    I remembered that bit of trivia . . . and I was literally LOL.

    Then I had to devise an approach . . . and I used Arithmetic Series.

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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, ecMathGeek!

    A straight-forward, no-nonsense approach . . . great job!

    I intended to use your method, but something bothered me.

    The square root of the sum? . . . Won't that be some ugly decimal?
    . . Why in the world would they want such a . . . Oh!

    I remembered that bit of trivia . . . and I was literally LOL.

    Then I had to devise an approach . . . and I used Arithmetic Series.

    Thank you.

    I started it thinking the same thing (but figured/hoped it would work out in the end). Just after I set up the summation, I remembered that same trivia that you spoke of and knew at that point what the answer would be. So, relieved that there would be a solution, I persevered until I reached it.
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