## anonymous 4 years ago find the sum ∑_(n=1)^100▒2n

1. anonymous

hmmm

2. Australopithecus

is this summation censored?

3. anonymous

no

4. anonymous

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

5. anonymous

yes

6. KingGeorge

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

7. KingGeorge

Where $$n=100$$

8. anonymous

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

9. anonymous

still not getting it

10. 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

11. anonymous

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

12. Australopithecus

100 terms sorry

13. anonymous

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

14. anonymous

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

15. Australopithecus

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

16. Australopithecus

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

17. Australopithecus

or you can just accept it

18. 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

19. 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

20. anonymous

okay so c=2 n=100

21. anonymous

yes

22. anonymous

okay got it thanks

23. anonymous

welcome