Question

Find the sum of each arithmetic series.

the first 100 even natural numbers.

Answer 1

@radar
@robtobey
@amistre64
@ivettef365

Answer 2

2+4+6+8+.....+100
200 +198 + 196+....+102
add top tp bottom
202+202+202...+202
there are 50 of these
so
50(202) =

Answer 3

guass already did this one lol

Answer 4

I think

Answer 5

gauss did 1 to 100 ... i forgot to actually comprehend the problem

Answer 6

@zzr0ck3r how did u get 202

Answer 7

im adding top to bottom

Answer 8

\[\sum_{n=1}^{100}~2n\]
\[2\sum_{n=1}^{100}~n\]
\[2\frac{100(101)}{2}\]

Answer 9

2 + 4 + 6 + ........................ + 100
200 + 198 + 196 + ..............................+ 102

200+2 = 202
198+4 = 202
196+6 = 202
......
we will have 50 of these
so
50(202) = 10100

Answer 10

@amistre64 what did you plug in for the variables in the sigma notation

Answer 11

the numbers 1 to 100; that would count off the first 100 even numbers

Answer 12

waht did u plug in for n?

Answer 13

\[\sum_{n=1}^{100}~2n\]
\[2\sum_{n=1}^{100}~n\]
\[2(1+2+3+4+...+100)\]
\[or~\frac{100(2(1)+2(100))}{2}=100(1+100)\]

Answer 14

the sum of n is a well rehearsed formula:

n, times the sum of the first and last terms; divided in half

Answer 15

Thank you so much!

Answer 16

youre welcome