Question

Summation

Answer 1

Could you explain how to work with a problem? Like how do you know what to do?

Answer 2

first notice that the sign is flipping and the terms are increasing by "2"
its arithmetic sequence kindof

Answer 3

Ok I see this

Answer 4

the alternating sign can be represented using :
\(\sum \limits_{k=1}^6(-1)^k\)

Answer 5

knw how to expand that sum ?

Answer 6

so if it goes + - + - it is represented by that equation?

Answer 7

yes but you should not memorize it like that
these will come in many places so u need to expand it and see what u get

Answer 8

you haven't answered my question, do u knw how to expand the given summation ?

Answer 9

Oh sorry. I do not. How would you go about expanding?

Answer 10

lets start with a simpler summation first

Answer 11

oks

Answer 12

\(\large \sum \limits_{k=1}^3 k\)

Answer 13

lets expand this and see what we get

Answer 14

\(\large \sum \limits_{k=1}^3 k = 1 + 2 + 3\)

Answer 15

^^it expands to that

Answer 16

let me give u another simpler summation, lets see if u wil be able to figure it out how to expand :)

Answer 17

\(\large \sum \limits_{k=1}^3 k^2 = ?\)

Answer 18

Alrights c:

Answer 19

Would it be the same thing k=1 + 2 + 3?

Answer 20

or does the k^2 affect the whole equation

Answer 21

because its like saying 1^2

Answer 22

\(\large \sum \limits_{k=1}^3 k^2 = 1^2 + 2^2 + 3^2\)

Answer 23

lets try another example

Answer 24

oh so it all becomes squared

Answer 25

Alrights

Answer 26

you simply plugin the \(k\) value each time and add...

Answer 27

next example :

\(\large \sum \limits_{k=1}^3 \dfrac{1}{k} = ?\)

Answer 28

k= 1 + 1/2 + 1/3?

Answer 29

You got it,
one more last example

Answer 30

hey dont say k = 1+1/2+1/3
its wrong to say that

Answer 31

k goes 1,2,3

Answer 32

\(\large \sum \limits_{k=1}^3 \dfrac{1}{k} = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3}\)

Answer 33

the summation equals that^^
not the k

Answer 34

Wait so what I am trying to find this time o.o I am confused

Answer 35

the k?

Answer 36

you're just expanding the summation, thats all

Answer 37

lets do one more example

Answer 38

\(\large \sum \limits_{k=1}^3 2k = ? \)

Answer 39

k= 2 + 4 + 6?

Answer 40

excellent !

Answer 41

let me give u one more

Answer 42

\(\large \sum \limits_{k=1}^6 2k = ?\)

Answer 43

k= 2 + 4 + 12?

Answer 44

Again, its not \(k\), why you're saying k = 2+4+12 ?

Answer 45

its the entire summation thing that wraps out to that sum

Answer 46

oh /.\ I just saw you do it at the beginning so I thought... *sigh*

Answer 47

if u want u can say :
sum = 2+4+....

Answer 48

btw its wrong. let me correct it

Answer 49

\(\large \sum \limits_{k=1}^6 2k = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + 2(6) \)

\(\large ~~~~~~~~~~~~~= 2+4+6+8+10+12 \)

Answer 50

Oh alirght I believe I know how to do this nows. Can i have another example

Answer 51

since \(k\) is going from \(1\) to \(6\), u get \(6\) terms in the sum

Answer 52

Yes I just saw that Aha :3

Answer 53

good :)

Answer 54

\(\large \sum \limits_{k=1}^6 2^k = ? \)

Answer 55

one moment

Answer 56

ok im back so. sum = 2^2+4^2+6^2+8^2+10^2+12^2

Answer 57

oh wow. I thought that said
dang it one moment

Answer 58

sum= 2 + 4 + 8 + 16 + 32 + 64

Answer 59

yes but can u show me one step before that ?

Answer 60

2^(1) 2^(2) 2^(3) 2^(4) 2^(5) 2^(6)

Answer 61

\(\large \sum \limits_{k=1}^6 2^k = 2^1 + 2^2 + 2^3 + 2^4+2^5+2^6\)
\(\large ~~~~~~~~~=2 + 4+ 8+16+32+64\)

Answer 62

Yes ! looks you got it completely !!

Answer 63

:D

Answer 64

one more last example before doing ur actual problem

Answer 65

I actually know the answer to my problem hue :3

Answer 66

and oks

Answer 67

\(\large \sum \limits_{k=1}^6 (-1)^k = ? \)

Answer 68

-1^1 -1^2 -1^3 -1^4 -1^5 -1^6 sum= -1 +1 -1 + 1 - 1 + 1

Answer 69

thats it !!

Answer 70

once u knw what they are, summations are very easy

Answer 71

yes aha :3

Answer 72

So for the actual problem would it be B?

Answer 73

nope expand it and see what u get ?

Answer 74

well for #1 -2 + 4 - 8 which is wrong

Answer 75

# 2 -2 + 4 - 8 again

Answer 76

#3 it starts with 4 which leaves us with #4

Answer 77

*sigh* i think i did this wrong.

Answer 78

Again : you should get \(6\) terms when u expand right ?

Answer 79

why are u showing only 4 terms ?

Answer 80

Because I saw that it would not match up with the numbers given

Answer 81

So I kinda was just cutting to the chase

Answer 82

option A :
\(\sum \limits_{k=1}^6 (-1)(2)^k = (-1)(2)^1 + (-1)(2)^2 + (-1)(2)^3 + (-1)(2)^4 + (-1)(2)^5 + (-1)(2)^6 \)
\(~~~~~~~~~~~~~~~~~= -2^1 - 2^2 - 2^3 - 2^4 - 2^5 - 2^6\)

Answer 83

for first option, u wud get all negative numbers^

Answer 84

so thats not the correct option

Answer 85

Oh... *facepalm* i worked out the exponents stupid me

Answer 86

show me the full work for option B
(no cutting to the chase :))

Answer 87

aha oks :3

Answer 88

(-2)^k= -2^1 -2^2 -2^3 -2^4 -2^5 -2^6

Answer 89

nope

Answer 90

dont eat parenthesis, show it as well

Answer 91

wait what did i do wrong

Answer 92

and where is the SUM notation ?

Answer 93

ur statement above is meaningless without the sum notation

Answer 94

OH
I didnt work in full

Answer 95

(-2)^k (-2)^1 (-2)^2 (-2)^3 (-2)^4 (-2)^5 (-2)^6

sum= -2^1 -2^2 -2^3 -2^4 -2^5 -2^6

Answer 96

?