Question

what is the sum of the terms of the series (-1)+(-2)+(-3)+...+(-10) help !!

Answer 1

well with just those four numbers it would be -16

Answer 2

-55

Answer 3

-1 + -2 + -3 + -4 + -5 + -6 + -7 + -8 + -9 + -10 = -55

Answer 4

oh i see so he was just posting it the faster way... @StudyGurl14 is right

Answer 5

its arithmetic series, and its actually 10 numbers.

Answer 6

give her a medal!

Answer 7

Also, you can use the following formula...

Answer 8

can you explain to me how you got -55?

Answer 9

where n = number of terms, a1 = first term, a2 = nth term, Sn= sum
\(\Large S_n=\frac{n(a_1+a_n)}{2}\)

Answer 10

\(\Large S_{10}=\frac{10(-1+-10)}{2}\rightarrow \frac{10(-11)}{2}\rightarrow\frac{-110}{2}=-55\)

Answer 11

Or you can factor -1 out, and you have \(-1(1+2+3+\cdots+10)\)

To find the sum fron 1 to n: you have \(S = \dfrac{n(n+1)}{2}\)

So you have \(-\dfrac{10(11)}{2} = -\dfrac{110}{2}=\boxed{-55}\)

Answer 12

True @geerky42

Answer 13

thanks for y'all help , i appreciate this a lot. i wish you guys can help me out a lot more with this .

Answer 14

:)

Answer 15

expand.
\[\sum_{n=2}^{6}3n\]

Answer 16

Just use formula @StudyGurl14 gave you. \(S_n = \dfrac{n(a_1+a_n)}{2}\)

For first term, n=2 so you have \(a_1=3(2)=6\)
For final term, n=6 so you have \(a_n = 3(6)=18\)
And there are \(6-2+1=5\)terms.

So you have \(\dfrac{5(6+18)}{2}=\boxed{60}\)

Answer 17

where did you get 6-2+1= 5 terms from? @geerky42

Answer 18

it says expand, its not asking for the sum.

Answer 19

Shortcut to count terms, you just subtract upper limit by lower limit, then add one.

Instead of count how many terms 2, 3, 4, 5, 6, you just do math: \(6-2+1=5\)

So we know that there are five terms. Useful if you have a ton of terms to count.

Answer 20

why are you adding 1 , where does that come from?

Answer 21

Well, expanding should be easy.

\[\sum_{n=2}^63n = 3(2)+3(3)+3(4)+3(5)+3(6)\]I am sure you can simplify this expansion yourself.


We add one because when we subtract upper limit by lower limit, we exclude lower limit itself, so we have to add one to make it counts.

See: how many term from 2 to 3? Two, right? just 2 and 3.

Try different approach: \(3-2=1\) So we need to add one.

Does that make sense?

Answer 22

I got the answer 60 too, but the expansion you gave above does not equal 60.... @geerky42

Answer 23

Check again.

Answer 24

Oops, sorry i hit the wrong button on my calculator. Yeah, it equals 60, lol.

Answer 25

Nice job @geerky42

Answer 26

2,3,4,5,6 wasn't the right answer for the expansion. so , i dont get it . this is very frustrating. @geerky42

Answer 27

"2,3,4,5,6" is not right way to answer...

I already gave you expansion. Now you just multiply number (don't add), then submit that.

Answer 28

The expansion was not 2,3,4,5,6.... @znae
It was
3(2) + 3(3) + 3(4) + 3(5) + 3(6)

Answer 29

oh ok, i see . thanks for the help y'all . @geerky42 @StudyGurl14

Answer 30

:)

Answer 31

one more . \[\sum_{n=1}^{6}(n-1)\]

Answer 32

What are you doing? Solving, or expanding?

Answer 33

expanding.

Answer 34

Do this:
(1 - 1) + (2 -1) + (3 - 1) + (4 - 1) + (5 - 1) + (6 - 1)
0 + 1 + 2 + 3 + 4 + 5

@geerky42 Do you know whether or not the 0 is included?

Answer 35

yes the 0 is included . thanks.