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teachme1234

  • 2 years ago

find the sum ∑_(n=1)^100▒2n

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  1. satellite73
    • 2 years ago
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    hmmm

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

  3. teachme1234
    • 2 years ago
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    no

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

  5. teachme1234
    • 2 years ago
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    yes

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

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

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

  9. teachme1234
    • 2 years ago
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    still not getting it

  10. Australopithecus
    • 2 years ago
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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

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

  12. Australopithecus
    • 2 years ago
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    100 terms sorry

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

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

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

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

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

  18. colorful
    • 2 years ago
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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

  19. colorful
    • 2 years ago
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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

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

  21. colorful
    • 2 years ago
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    yes

  22. teachme1234
    • 2 years ago
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    okay got it thanks

  23. colorful
    • 2 years ago
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    welcome

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