Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

AustralopithecusBest ResponseYou've already chosen the best response.0
is this summation censored?
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
\[\sum_{n=1}^{100}2n\]?
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
\[\sum_{k=1}^nk=\frac{n(n+1)}{2}\] so ... what king george said
 one year ago

teachme1234Best ResponseYou've already chosen the best response.0
still not getting it
 one year ago

AustralopithecusBest 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
 one year ago

teachme1234Best ResponseYou've already chosen the best response.0
we are counting from 1 to 100 what is the nth term.
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
100 terms sorry
 one year ago

teachme1234Best ResponseYou've already chosen the best response.0
2xn(n+1)/2 so i use this fromula to solve it.
 one year ago

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

AustralopithecusBest 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
 one year ago

AustralopithecusBest 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
 one year ago

AustralopithecusBest ResponseYou've already chosen the best response.0
or you can just accept it
 one year ago

colorfulBest 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
 one year ago

colorfulBest 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
 one year ago

teachme1234Best ResponseYou've already chosen the best response.0
okay got it thanks
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.