find the sum ∑_(n=1)^100▒2n

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find the sum ∑_(n=1)^100▒2n

Mathematics
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hmmm
is this summation censored?
no

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Other answers:

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

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