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Loser66

  • one year ago

We have \[\sum_{j =1}^n j = \dfrac{n(n+1)}{2}\] How about the limit from m to n? I meant \[\sum_{j =m}^n j=??\] Please, help

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  1. ParthKohli
    • one year ago
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    \[= \sum_{1}^{n}j - \sum_{1}^{m-1}j\]

  2. ParthKohli
    • one year ago
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    sorry for lack of notation

  3. imqwerty
    • one year ago
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    terms from m to n = n-m+1 so summation wuld be- \[\frac{ (n-m+1)(m+n) }{ 2 }\]

  4. imqwerty
    • one year ago
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    number of terms =n-m+1 1st term =m and common diff=1 so just put this in the formula of summation of ap :)

  5. Loser66
    • one year ago
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    Got it. thank you.

  6. imqwerty
    • one year ago
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    np :)

  7. Loser66
    • one year ago
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    In general, if we see the lower limit is not 1, when we see n, we replace by what? hehehe. I am silly, just want to generalize the formula like \(\sum_{j = 1}^n = \dfrac{n(n+1)}{2} \) the second n+ 1 is upper + lower limit, right? the first n is upper - lower +1 Hence for all lower limit, we just apply it, right?

  8. Loser66
    • one year ago
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    |dw:1443637998925:dw|

  9. Loser66
    • one year ago
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    |dw:1443638041475:dw|

  10. ParthKohli
    • one year ago
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    can't you apply that in the same way?

  11. Loser66
    • one year ago
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    I don't know, I ask and need a confirmation or a logic how to generalize it. Since I don't think we can.

  12. ParthKohli
    • one year ago
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    \[\sum_{j=m}^{ n}f(j) = \sum_{j=1}^n f(j) - \sum_{j = 1}^{m-1}f(j)\]Do you see why the above is true?

  13. Loser66
    • one year ago
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    Yes,

  14. ParthKohli
    • one year ago
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    Cool. So it can be generalised, right?

  15. Loser66
    • one year ago
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    yes

  16. ParthKohli
    • one year ago
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    Great.

  17. ParthKohli
    • one year ago
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    Just wondering - are you an Indian?

  18. Loser66
    • one year ago
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    ok, got it. the first sum = (n(n+1) )/2 the second sum = ((m-1)(m-1+1))/2 some algebraic calculation and we get the answer. Thanks a lot No, I am not. I am Vietnamese

  19. imqwerty
    • one year ago
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    sry there was some server update i had to go >.<

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