Question

Imaginary number question.. Imaginary number question.. @Mathematics

Answer 1

\[(i \times i^{4} \times i^{7} \times i^{10} \times i^{13} \times ... \times i^{58} \times i^{61} \times i^{64}) \div (i + i^{4} + i^{7} + i^{10} + i^{13} + ... + i^{58} + i^{61} + i^{64})\]

Answer 2

erm
how do i put it as a fraction?

Answer 3

O.o

Answer 4

1?

Answer 5

\[\frac{i \times i^{4} \times i^{7} \times i^{10} \times i^{13} \times ... \times i^{58} \times i^{61} \times i^{64}}{i + i^{4} + i^{7} + i^{10} + i^{13} + ... + i^{58} + i^{61} + i^{64}}\]

Answer 6

nvm it's plus on the bottom lol

Answer 7

thank you myininaya :D

Answer 8

\[\prod_{k=0}^{21}i ^{1+3k} \div \sum_{k=0}^{21}i ^{1+3k} \]

Answer 9

O_o, can I compute that by hand? Because I can't use a calculator

Answer 10

\[i+i^4+i^7+\cdots+i^{64}=\frac{i^{67}-i}{i-1}=\frac{-2i}{i-1}\]by geometric formula.

Answer 11

Well, the sum of the exponents on the top will yield an odd number...

Answer 12

You can probalbly find a pattern for the multiplication
and for the addition part, use the geometric formula

Answer 13

http://www.wolframalpha.com/input/?i=%5Cprod_%7Bk%3D0%7D%5E%7B21%7Di+%5E%7B1%2B3k%7D+%5Cdiv+%5Csum_%7Bk%3D0%7D%5E%7B21%7Di+%5E%7B1%2B3k%7D

Answer 14

oh, and the fact that:\[i^{67}=i^3=-i\]

Answer 15

\[\frac{i^{1+4+7+13+ \cdot \cdot \cdot 64}}{i+1-i-1+i+ \cdot \cdot \cdot -1+i+1}\]
it looks like a pattern on bottom

Answer 16

\[\frac{2i}{1-i}=\frac{2i(1+i)}{2}=-1+i\]this is the denominator.

Answer 17

how do you get\[ \frac{2i}{1-i}\]

Answer 18

From what myininaya said,\[\frac{i^{715}}{i+1}\]

Answer 19

did you see how i got:\[\frac{-2i}{i-1}\]? its just multiplying by -1 on top and bottom.

Answer 20

\[\frac{i^{\sum_{k=1}^{21}(3k+1)}}{-1+i}\]
---
\[\sum_{k=1}^{21}3k+\sum_{k=1}^{21}1=3 \cdot \frac{21(21+1)}{2}+21(1)\]

Answer 21

\[=693+21=714\]

Answer 22

now we need to divide this number by 4 to see what the remainder is

Answer 23

i = sqrt(-1)
i^2 = -1
i^3 = -sqrt(-1)
i^4 = 1
1+4+7...64 = 22*65/2=715
715/4 = 178 + 3/4 = -i
a = i
r = -i
n = 21

\[\frac{i((-i)^{21}-1}{-i-1}\]

Answer 24

remainder is 2

Answer 25

i^2=-1

Answer 26

\[\frac{-1}{-1+i}\]

Answer 27

i think it's 715 not 714

Answer 28

it is 715

Answer 29

Ok I get how the numerator is -i, but how do I get the denominator?

Answer 30

oh i missed a number

Answer 31

i forgot to include 1 in my summation

Answer 32

i started with the 4

Answer 33

the denominator is in my posts, geometric formula.

Answer 34

\[i+i^4+i^7+\cdots+i^{61}+i^{64}=\frac{i^{67}-i}{i-1}\]

Answer 35

i^67 is i^3 which is -i, so we get:

Answer 36

@joemath314159
It should not be 67, since it does not have 67 terms

Answer 37

\[\frac{-i-i}{i-1}=\frac{-2i}{i-1}\]

Answer 38

\[715=4(178)+3=>i^{715}=i^{3}=-i\]

Answer 39

you can't do that joe

Answer 40

joe got in trouble

Answer 41

lol

Answer 42

i did something wrong >.<

Answer 43

tn = i(i)^(n-1)
64 = i(i)^(n-1)

Answer 44

okay so the denominator is the geometric progression \[i(i^3)^{n-1}\] right? so what do i do from there

Answer 45

\[\sum_{n=0}^{21}i^{3n+1}\]

\[=i\sum_{n=0}^{21}(i^3)^{n}=i\frac{(i^3)^{22}-1}{i^3-1}=1+i\]

Answer 46

i think this might be the most people who tried to help one person at the same time

Answer 47

http://www.wolframalpha.com/input/?i=%5Cprod_%7Bk%3D0%7D%5E%7B21%7Di+%5E%7B1%2B3k%7D+%5Cdiv+%5Csum_%7Bk%3D0%7D%5E%7B21%7Di+%5E%7B1%2B3k%7D

Answer 48

i got the denominator wrong -.- but you can do it this way.

Answer 49

\[\frac{-i}{1+i}=\frac{-1}{2}-\frac{i}{2}\]

Answer 50

\[\frac{-i}{1+i} \cdot \frac{1-i}{1-i}=-\frac{i-i^2}{1-i^2}=-\frac{i-(-1)}{1-(-1)}=-\frac{i+1}{2}\]

Answer 51

Yea that's probably the answer

Answer 52

zarkon it looks prettier written as one fraction

Answer 53

it does

Answer 54

:)

Answer 55

is this a college algebra problem for real?

Answer 56

the answer is.. \[-\frac{1}{2}-\frac{1}{2}i\] so woooh thank you guys :D

Answer 57

are you in college algebra?

Answer 58

Glad that we can help

Answer 59

and nope, this is a math team question haha

Answer 60

oh what does that mean

Answer 61

Math contest?

Answer 62

math team..
i don't have a math team

Answer 63

it's like a competition between high schools. schools do practice problems and then compete

Answer 64

Like math counts

Answer 65

thats cool
i wish my school did something like that
or participated in something like that

Answer 66

Texas has something called UIL that does high school math competitions.

Answer 67

we would lose those since my school sucked!

Answer 68

I did AMC before

Answer 69

My school is 43rd in the state haha

Answer 70

i bet my school would be like 5000th place

Answer 71

how many schools are there
that might be too good for my high school lol

Answer 72

There's the Putnam exam for undergraduates. thats fun stuff.

Answer 73

i remember the putnam exam i took it once
i did it for bonus points lol
that junk was kindof hard for me

Answer 74

i was still getting my bachelors though

Answer 75

maybe i got a tiny bit smarter who knows

Answer 76

well not smarter
but more creative

Answer 77

math is like art right?

Answer 78

art of problem solving

Answer 79

indeed, a good proof is a work of art :)