Find the sum of each arithmetic series.

the first 100 even natural numbers.

Question

Find the sum of each arithmetic series.

the first 100 even natural numbers.

anonymous · Answer

@radar 
@robtobey 
@amistre64 
@ivettef365

zzr0ck3r · Answer

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) =

amistre64 · Answer

guass already did this one lol

zzr0ck3r · Answer

I think

amistre64 · Answer

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

anonymous · Answer

@zzr0ck3r  how did u get 202

zzr0ck3r · Answer

im adding top to bottom

amistre64 · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

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

amistre64 · Answer

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

anonymous · Answer

waht did u plug in for n?

amistre64 · Answer

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

amistre64 · Answer

the sum of n is a well rehearsed formula:

n, times the sum of the first and last terms;  divided in half

anonymous · Answer

Thank you so much!

amistre64 · Answer

youre welcome