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Curry

  • 5 years ago

sigma

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  1. Curry
    • 5 years ago
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    \[\sum_{x=1}^{25}\] K^2 -3k +4 what is the formula to solve if u dont have calculator

  2. Curry
    • 5 years ago
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    please help

  3. anonymous
    • 5 years ago
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    \[\sum_{x=1}^{25}k^2-3\sum_{x=1}^{25}k+4\sum_{x=1}^{25}1\]

  4. anonymous
    • 5 years ago
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    imran please go ahead, i have to get back to work

  5. anonymous
    • 5 years ago
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    but this one is pretty easy \[4\sum_{k=1}^{25}1=4\times 25=100\]

  6. anonymous
    • 5 years ago
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    sum 1 to n of (1) is equal to n sum 1 to n of (i) is equal to n(n+1)/2 sum 1 to n of (i^2) is equal to n(n+1)(2n+1)/6 um i think.

  7. anonymous
    • 5 years ago
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    yes, agreed^

  8. anonymous
    • 5 years ago
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    @curry you got this?

  9. anonymous
    • 5 years ago
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    \[\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\] so \[-3\sum_{k=1}^{25}k=-3\times \frac{25\times 26}{2}\]

  10. Curry
    • 5 years ago
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    kind a

  11. anonymous
    • 5 years ago
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    \[\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}\] so \[\sum_{k=1}^{25}k^2=\frac{25(26)(51)}{6}\]

  12. anonymous
    • 5 years ago
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    now you have all the numbers you need, grind it out with a calculator

  13. anonymous
    • 5 years ago
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    should get 4650 if you do it right http://www.wolframalpha.com/input/?i=sum+k+%3D+1+to+25+%28k^2-3k%2B4%29

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