Question

HELLLPPPPPPP!!!!! find the 8th term:

Answer 1

\[(2x-4)^{13}\]

Answer 2

have you covered the "binomial theorem" yet?

Answer 3

http://www.youtube.com/watch?v=1pSD8cYyqUo

Answer 4

i took the class last year but i never took the final exam so my teacher gave me until spring to finish it. i did go over that theorem but last year so i do not remember..

Answer 5

\[Tr+1=\left(\begin{matrix}n \\r\end{matrix}\right)y ^{n-r}a ^{r}\in\left( y+a \right)^{n}\]
put y=2x,a=-4,n=13,r+1=8

Answer 6

woahhhh....wutt????

Answer 7

@RadEn pleaseee helppp

Answer 8

well, the 8th term of (2x -4)^13 is
13C7 (2x)^(13-7) * (-4)^7

Answer 9

what? it makes no sense to me

Answer 10

there are 3 cases here :

13C7 = ... ?

(2x)^6 = ... ?

(-4)^7 = ... ?

Answer 11

whats C?

Answer 12

C = combination symbol

use this formula to get 13C7
nCr = n!/(r!(n-r)!)

Answer 13

\(\bf (2x-4)^{13}\\ \quad \\ \quad \\
(2x)^{13}+13(2x)^{12}(-4)^1+\square (2x)^{11}(-4)^2+\square (2x)^{10}(-4)^3\\ \quad \\+\square (2x)^{9}(-4)^4+
\square (2x)^8(-4)^5+\square (2x)^7(-4)^6...\)

notice as the 1st term exponent DECREASES, the 2nd term exponent INCREASES simultaneously, that's pretty straightforward
now to find the coefficients, is the only part that you'd need

Answer 14

ok look fir this question i had an answer written down but im not sure if its right

i wrote \[12^{13}=1.069932 \times 10^{14}\]

Answer 15

hmmm... well... if you dunno the binomial theorem... how did you get that?

Answer 16

ohh that's another one?

Answer 17

i dont know i think i asked it before but i dont remember it al all so i wanted to make sure it was right

Answer 18

\(\bf 12^{13}=106993205379072\implies 106993205379072\times 10^1\\ \quad \\
\implies 1.06993205379072\times 10^{14}\)

why? well, there are 15 digits, you move the decimal to the left 14 spots, and multiply by a likewise base 10, it'd equate, yes

Answer 19

so it is in fact correct...do you by any chance know how i got it.... lol

Answer 20

this site might useful.
http://www.mathsisfun.com/algebra/binomial-theorem.html

Answer 21

I wonder that myself =)

keep in mind that exercises aren't there for answers to be just transcribed over, they're only useful if you do them, no transcribe material over

Answer 22

I JUST FIGURED OUT HOW I GOT IT!!!!

Answer 23

good

Answer 24

i did.... so if im looking for the 8th term i plugged in 8 for x so i did \[(2(8)-4)^{13}\] so then i got \[12^{13}\] and that is what i put

Answer 25

heheh

Answer 26

setting x = 8, is not the same as the 8th term in the sequence
I said your material of \(\bf 12^{13}=106993205379072\) is correct, you never mentioned it was your 8th term, so, no it's not the 8th term
x = 8 and the 8th term aren't the same thing

check the video above or the urls posted for the binomial theorem... :) is a rather short video you know

Answer 27

can you just give me the answer.....? haha cuz i really dont get it

Answer 28

it's not easy to get it when not reading the material
so... check the video or urls

Answer 29

ok thanks...............