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

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1. satellite73 Group Title

hmmm

2. Australopithecus Group Title

is this summation censored?

3. teachme1234 Group Title

no

4. satellite73 Group Title

$\sum_{n=1}^{100}2n$?

5. teachme1234 Group Title

yes

6. KingGeorge Group Title

$\sum_1^{100} 2n=2 \cdot \sum_1^{100} n=2\cdot {n(n+1) \over 2}$

7. KingGeorge Group Title

Where $$n=100$$

8. satellite73 Group Title

$\sum_{k=1}^nk=\frac{n(n+1)}{2}$ so ... what king george said

9. teachme1234 Group Title

still not getting it

10. Australopithecus Group Title

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 Group Title

we are counting from 1 to 100 what is the nth term.

12. Australopithecus Group Title

100 terms sorry

13. teachme1234 Group Title

2xn(n+1)/2 so i use this fromula to solve it.

14. colorful Group Title

in general $\sum_{i=1}^{n} i={n(n+1)\over2}$

15. Australopithecus Group Title

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

16. Australopithecus Group Title

if you want to understand the formula I recommend just looking up sum of arithmetic formula on wiki

17. Australopithecus Group Title

or you can just accept it

18. colorful Group Title

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 Group Title

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 Group Title

okay so c=2 n=100

21. colorful Group Title

yes

22. teachme1234 Group Title

okay got it thanks

23. colorful Group Title

welcome