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Math Help - "Rewrite using summation notation"

  1. #1
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    "Rewrite using summation notation"

    I understand how to do summation notation and the formula for it (n(n+1))/2 and how that differs if n has an exponent in it, but what if you're asked to rewrite a problem in summation notation?

    For example "Rewrite 1+2+3...+100 in summation notation"
    and just so I have a clear understanding let's also say
    "Rewrite 1^2+2^2+3^2...+100^2 in summation notation"

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by fattydq View Post
    I understand how to do summation notation and the formula for it (n(n+1))/2 and how that differs if n has an exponent in it, but what if you're asked to rewrite a problem in summation notation?

    For example "Rewrite 1+2+3...+100 in summation notation"
    and just so I have a clear understanding let's also say
    "Rewrite 1^2+2^2+3^2...+100^2 in summation notation"

    Thanks in advance!
    For the first one, let i be your counter,
    so it would be \sum_{i=1}^{100}i

    For the second one, still let i be your counter,
    so it would be \sum_{i=1}^{100}i^2
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  3. #3
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    Quote Originally Posted by chabmgph View Post
    For the first one, let i be your counter,
    so it would be \sum_{i=1}^{100}i

    For the second one, still let i be your counter,
    so it would be \sum_{i=1}^{100}i^2

    oh wow, that's really all they mean you think? And what does the i=1 thing mean, is that the interval between the values or something? So if it went 1+3+5+7 i would equal 2?
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  4. #4
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    Quote Originally Posted by fattydq View Post
    oh wow, that's really all they mean you think? And what does the i=1 thing mean, is that the interval between the values or something? So if it went 1+3+5+7 i would equal 2?
    Your right, the common jump is two. If you write the terms like

    <br />
\begin{array}{c}<br />
(1+2\cdot 0) + (1+2\cdot 1) + (1+2\cdot 2) + (1 + 2\cdot 3) + \cdots<br />
\end{array}

    then

    \sum_{i=0}^4 1 + 2i.
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  5. #5
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    Quote Originally Posted by danny arrigo View Post
    Your right, the common jump is two. If you write the terms like

    <br />
\begin{array}{c}<br />
(1+2\cdot 0) + (1+2\cdot 1) + (1+2\cdot 2) + (1 + 2\cdot 3) + \cdots<br />
\end{array}

    then

    \sum_{i=0}^4 1 + 2i.

    gotcha, thanks a bunch man
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