Question

Find, in simplest form, the coefficient of x^r in the expansion, in ascending powers of x, of [1/(1-x)^3]

Answer 1

(1-x)^-3

Answer 2

use binomial theorem

(1 + (-x))^-3 =

inf
∑(-3Ck) (-x)^n
k=0

now it's just the matter of expanding (-3Ck)

Answer 3

typo, should be

inf
∑(-3Cn) (-x)^n
n=0

Answer 4

can you take it from here?

Answer 5

sorry i dont understand the ∑(-3Cn) (-x)^n,
im suppose to use the [(n)(n-1)..(n-r+1)]/r! formula

Answer 6

yes, take's exactly what you use.

(-3Cn) = (-3)(-4)(-5) ... (-3-n+1) / n!
= (-3)(-4)(-5) ... (-2-n) / n!
= (-1)^n (3*4*5 * ... * (n+2) ) / n!
= (-1)^n (1*2*3*4*5 *... * n (n+1) (n+2) ) /(1*2* n!)
= (-1)^n (1/2) (n+1) (n+2)

Answer 7

huhm.. seems like i'm off by a constant 1/2. Hold on, let me check my algebra again

Answer 8

yes im self-studying

Answer 9

@HatcrewS are you trying to find the power expansion form?

Answer 10

im trying to find the general term for the coefficient of the expansion, the answer is
[(r+1)(r+2)]/2

Answer 11

which is exactly what i got O.O

Answer 12

hmm but you got the (-1)^n >_<
and i still don't understand how you get (-1)^n, how does the ^n come about ?
I just memorise that if the sequence is negative I can take out -1 and make it become (-1)^n

Answer 13

you can eliminate the (-1)^n by having another (-1)^n from (-x), then somehow making it (-1)^2n, which is always positive i guess :/

Answer 14

that is because if you multiply a negative odd number of times, you'll get a negative number. And if you multiply a negative number an even number of times, then it's positive.

(-1)^n determines the sign of the coefficient

Answer 15

ah yes, (-1)^n * (-1)^n at the end will always end up positive. I didn't simplify all the way through