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teachme1234
 2 years ago
find the sum ∑_(n=1)^100▒2n
teachme1234
 2 years ago
find the sum ∑_(n=1)^100▒2n

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

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

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

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

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

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0you 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

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

Australopithecus
 2 years ago
Best ResponseYou've already chosen the best response.0100 terms sorry

teachme1234
 2 years ago
Best ResponseYou've already chosen the best response.02xn(n+1)/2 so i use this fromula to solve it.

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

Australopithecus
 2 years ago
Best 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

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

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

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0the 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

colorful
 2 years ago
Best ResponseYou've already chosen the best response.0The 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
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