anonymous
  • anonymous
find the sum ∑_(n=1)^100▒2n
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
hmmm
Australopithecus
  • Australopithecus
is this summation censored?
anonymous
  • anonymous
no

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anonymous
  • anonymous
\[\sum_{n=1}^{100}2n\]?
anonymous
  • anonymous
yes
KingGeorge
  • KingGeorge
\[\sum_1^{100} 2n=2 \cdot \sum_1^{100} n=2\cdot {n(n+1) \over 2}\]
KingGeorge
  • KingGeorge
Where \(n=100\)
anonymous
  • anonymous
\[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] so ... what king george said
anonymous
  • anonymous
still not getting it
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
we are counting from 1 to 100 what is the nth term.
Australopithecus
  • Australopithecus
100 terms sorry
anonymous
  • anonymous
2xn(n+1)/2 so i use this fromula to solve it.
anonymous
  • anonymous
in general \[\sum_{i=1}^{n} i={n(n+1)\over2}\]
Australopithecus
  • Australopithecus
(2)100(100+1)/2 = 10100 proof https://www.wolframalpha.com/input/?i=summation+of+1+to+100+of+2n
Australopithecus
  • Australopithecus
if you want to understand the formula I recommend just looking up sum of arithmetic formula on wiki
Australopithecus
  • Australopithecus
or you can just accept it
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
okay so c=2 n=100
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay got it thanks
anonymous
  • anonymous
welcome

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