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Math Help - help finding sum of series

  1. #1
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    Exclamation help finding sum of series

    got 2 series.

    1.

    12 - 22 + 32 - 42 + ................. 992 - 1002

    (ok those are just squares)

    2.

    1.1! + 2.2! + ....... 50.50!

    need to find sum...
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  2. #2
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    Quote Originally Posted by adhyeta View Post
    got 2 series.

    1.

    12 - 22 + 32 - 42 + ................. 992 - 1002

    (ok those are just squares)



    need to find sum...
    1^2-2^2+3^2-4^2+\cdots+99^2-100^2

    1-4+9-16+25-36+\cdots+99^2-100^2

    finding the difference in pairs now giving 50 terms.

    -3-7-11-\cdots

    Use sum of an arithmetic sequence

    S_n = \frac{n}{2}(2a+(n-1)d)

    where n = number of terms = 50

    a = the first term = -3 and d = the common difference between terms = -4

    S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots
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  3. #3
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    Hello, adhyeta!

    Are you familiar with these summation formulas?

    . . \sum^n_{k=1} 1 \;\;= \;n

    . . \sum^n_{k=1}k \;\;=\;\frac{n(n+1)}{2}


    1)\;\;S \;=\;1^2 - 2^2 + 3^2 - 4^2 + \hdots + 99^2 - 100^2

    \text{We have: }\;S \;=\;\underbrace{(1^2 + 3^2 + 5^2 + \hdots + 99^2)}_{\text{first 50 odd squares}} \;-\; \underbrace{(2^2+4^2+6^2 + \hdots + 100^2)}_{\text{first 50 even squares}}

    . . \text{Then: }S \;=\qquad\qquad \sum^{50}_{k=1}(2k-1)^2 \qquad - \qquad\qquad \sum^{50}_{k=1}(2k)^2


    Hence: . S \;=\;\sum^{50}_{k=1}\bigg[(2k-1)^2 - (2k)^2\bigg] \;=\;\sum^{50}_{k=1}(1-4k)

    . . = \;\;\sum^{50}_{k=1}\!1 \;-\; 4\!\sum^{50}_{k=1}k \;\;=\;\;50 - 4\,\frac{50\cdot51}{2} \;\;=\;\;\boxed{-5,\!050}

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  4. #4
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    Quote Originally Posted by pickslides View Post
    1^2-2^2+3^2-4^2+\cdots+99^2-100^2

    1-4+9-16+25-36+\cdots+99^2-100^2

    finding the difference in pairs now giving 50 terms.

    -3-7-11-\cdots

    Use sum of an arithmetic sequence

    S_n = \frac{n}{2}(2a+(n-1)d)

    where n = number of terms = 50

    a = the first term = -3 and d = the common difference between terms = -4

    S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4)) = \cdots
    thats not coming out to be correct...its -5050.
    refer to soroban's solution.
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  5. #5
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    Quote Originally Posted by adhyeta View Post
    thats not coming out to be correct...its -5050.
    refer to soroban's solution.

    <br />
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))


    <br /> <br /> <br />
S_{50} = 25(-6+(49)(-4))


    <br /> <br /> <br />
S_{50} = 25(-6+-196)


    <br /> <br /> <br />
S_{50} = 25(-202)


    <br /> <br /> <br />
S_{50} = -5050


    Worked for me!
    Last edited by pickslides; April 30th 2009 at 02:16 PM. Reason: Typo
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  6. #6
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    Quote Originally Posted by adhyeta View Post
    got 2 series.

    1.

    12 - 22 + 32 - 42 + ................. 992 - 1002

    (ok those are just squares)

    2.

    1.1! + 2.2! + ....... 50.50!

    need to find sum...
    Here's the second one. Let

    S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!
     = (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!<br />
    = (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)
     <br />
= (n+1)! - 1<br />

    so here n = 50 so the answer S = 51!-1.
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  7. #7
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    Quote Originally Posted by pickslides View Post
    <br />
S_{50} = \frac{50}{2}(2(-3)+(50-1)(-4))


    <br /> <br /> <br />
S_{50} = 25(-6+(49)(-4))


    <br /> <br /> <br />
S_{50} = 25(-6+-196)


    <br /> <br /> <br />
S_{50} = 25(-202)


    <br /> <br /> <br />
S_{50} = -5050


    Worked for me!
    hey! so sorry. i think i went wrong with the calc.(-calculation-)
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  8. #8
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    Quote Originally Posted by danny arrigo View Post
    Here's the second one. Let

    S_n = 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \cdots + (n-1) \cdot (n-1)! + n \cdot n!
     = (2-1) 1! + (3-1)2! + (4-1)3! + \cdots + (n -1) \cdot (n-1)! + (n+1 -1) \cdot n!<br />
    = (2! - 1!) + (3! - 2!) + (4! - 3!) + \cdots + (n! - (n-1)!) + ((n+1)! - n!)
     <br />
= (n+1)! - 1<br />

    so here n = 50 so the answer S = 51!-1.
    hey! can we similarily prove the sum n^{2}.n!???
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