At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

hmmm

is this summation censored?

no

\[\sum_{n=1}^{100}2n\]?

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

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

okay so c=2 n=100

yes

okay got it thanks

welcome