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teachme1234

find the sum ∑_(n=1)^100▒2n

  • 2 years ago
  • 2 years ago

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  1. satellite73
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    hmmm

    • 2 years ago
  2. Australopithecus
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    is this summation censored?

    • 2 years ago
  3. teachme1234
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    no

    • 2 years ago
  4. satellite73
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    \[\sum_{n=1}^{100}2n\]?

    • 2 years ago
  5. teachme1234
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    yes

    • 2 years ago
  6. KingGeorge
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    \[\sum_1^{100} 2n=2 \cdot \sum_1^{100} n=2\cdot {n(n+1) \over 2}\]

    • 2 years ago
  7. KingGeorge
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    Where \(n=100\)

    • 2 years ago
  8. satellite73
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    \[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] so ... what king george said

    • 2 years ago
  9. teachme1234
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    still not getting it

    • 2 years ago
  10. Australopithecus
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    you can add them all up like 2(1) + 2(2) + 2(3)... or you can just use a formula to solve for it which is much quicker

    • 2 years ago
  11. teachme1234
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    we are counting from 1 to 100 what is the nth term.

    • 2 years ago
  12. Australopithecus
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    100 terms sorry

    • 2 years ago
  13. teachme1234
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    2xn(n+1)/2 so i use this fromula to solve it.

    • 2 years ago
  14. colorful
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    in general \[\sum_{i=1}^{n} i={n(n+1)\over2}\]

    • 2 years ago
  15. Australopithecus
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    (2)100(100+1)/2 = 10100 proof https://www.wolframalpha.com/input/?i=summation+of+1+to+100+of+2n

    • 2 years ago
  16. Australopithecus
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    if you want to understand the formula I recommend just looking up sum of arithmetic formula on wiki

    • 2 years ago
  17. Australopithecus
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    or you can just accept it

    • 2 years ago
  18. colorful
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    the rules of summations are such that we can take constants out of the summation sign\[\sum_{i=1}^{n} Ci=C\sum_{i=1}^{n} i=C{n(n+1)\over2}\]where C is any constant that is how we got our formula

    • 2 years ago
  19. colorful
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    The proof is quite easy if you know mathematical induction if not, it's better to just accept it, or look up the formula for any arithmetic series the wiki way, as suggested by @Australopithecus

    • 2 years ago
  20. teachme1234
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    okay so c=2 n=100

    • 2 years ago
  21. colorful
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    yes

    • 2 years ago
  22. teachme1234
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    okay got it thanks

    • 2 years ago
  23. colorful
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    welcome

    • 2 years ago
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