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Math Help - Sum of the squared of the integers

  1. #1
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    Sum of the squared of the integers

    Here is another section I am having trouble with, it asks for:
    the sum of the squares of the integers n: -12 less than or equal to n less than or equal 12.

    the sum of the reciprocals of the integers n, where 100 less than or equal to n than or equal 200.

    the sum of the cubes of the reciprocals of all integers, n where 1 is less than or equal to 50.

    Am I supposed to put this in sigma notation or in a sum of a series?

    I would appreciate it if you coud point me in the right direction.
    Thank you!
    Keith Stevens
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  2. #2
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    Hello, Keith!

    What were the instructions?
    . . "Write as a sum of a series" or "Write in sigma notation" ?
    I'll do it both ways . . .


    The sum of the squares of the integers for -12 \leq n \leq 12.
    (\text{-}12)^2 + (\text{-}11)^2 + (\text{-}10)^2 + \cdots + 10^2 + 11^2 + 12^2 \;=\;\sum^{12}_{n=-12} n^2


    The sum of the reciprocals of the integers for 100 \leq n \leq 200.
    \frac{1}{100} + \frac{1}{101} + \frac{1}{102} + \cdots + \frac{1}{200} \;=\;\sum^{200}_{n=100}\frac{1}{n}


    The sum of the cubes of the reciprocals for  1 \leq n \leq 50.
    \left(\frac{1}{1}\right)^3 + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{3}\right)^3 + \cdots + \left(\frac{1}{50}\right)^3\;=\;\sum^{50}_{n=1}\le  ft(\frac{1}{n}\right)^3

    Simplified: . \frac{1}{1^3} + \frac{1}{2^3} + \frac{1}{3^3} + \cdots + \frac{1}{50^3} \;=\;\sum^{50}_{n=1}\frac{1}{n^3}

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  3. #3
    Junior Member
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    Smile sum of squares

    Thos examples are exactly what I need to finish the rest of the problems!
    I am glad you can read my mind, sorry for not being more specific, I appreciate your insight.
    Thank you again!!!!!!!!!
    Keith Stevens
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