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Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
is this summation censored?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{n=1}^{100}2n\]?
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
\[\sum_1^{100} 2n=2 \cdot \sum_1^{100} n=2\cdot {n(n+1) \over 2}\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Where \(n=100\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] so ... what king george said
 2 years ago

teachme1234 Group TitleBest ResponseYou've already chosen the best response.0
still not getting it
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
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

teachme1234 Group TitleBest ResponseYou've already chosen the best response.0
we are counting from 1 to 100 what is the nth term.
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
100 terms sorry
 2 years ago

teachme1234 Group TitleBest ResponseYou've already chosen the best response.0
2xn(n+1)/2 so i use this fromula to solve it.
 2 years ago

colorful Group TitleBest ResponseYou've already chosen the best response.0
in general \[\sum_{i=1}^{n} i={n(n+1)\over2}\]
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
(2)100(100+1)/2 = 10100 proof https://www.wolframalpha.com/input/?i=summation+of+1+to+100+of+2n
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
if you want to understand the formula I recommend just looking up sum of arithmetic formula on wiki
 2 years ago

Australopithecus Group TitleBest ResponseYou've already chosen the best response.0
or you can just accept it
 2 years ago

colorful Group TitleBest ResponseYou've already chosen the best response.0
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

colorful Group TitleBest ResponseYou've already chosen the best response.0
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

teachme1234 Group TitleBest ResponseYou've already chosen the best response.0
okay so c=2 n=100
 2 years ago

teachme1234 Group TitleBest ResponseYou've already chosen the best response.0
okay got it thanks
 2 years ago
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