what is the macclaurin series for 1/(1-z)^3?

Question

PLEASE FOR THE LOVE OF GOD WHAT IS IT

I THINK MY ANSWER GUIDE IS WRONG!

AAAAAAAAHHHHHHHHHH I NEED TO KNOW BEFORE MY ENTIRE WORLD DISINTEGRATES

Answer 1

F(Z) = (1-Z)^-3, F(0) = 1
F'(Z) = 3(1-Z)^-4, F'(0) = 3
F''(Z) = 3*4(1-Z)^-5, F''(0) = 3*4
F'''(0) = 3*4*5, F''''(0) = 3*4*5*6
F(Z) = 1 + 3Z + 3*4(Z^2)/2! + 3*4*5(Z^3)/3! + ....
F(Z) = 1 + 3Z + 3*4(Z^2)/2 + 4*5(Z^3)/2 + 5*6(Z^4)/2 +....
F(Z) = 1/2[1*2 +2*3Z + 3*4*Z^2 + 4*5*Z^3 + 5*6*Z^4 + ...]

Answer 2

1/(1-z)^3 = (1-z)^-3=

1+(-3)(-z) +(-3)(-4)(-z^2)/2!+ +(-3)(-4)(-5)(-z)^3/3! +..+ a(a-1)..(a-n)(-z)^(n+1)/(n+1)! +++ where a in this case is(-3)

Answer 3

Sigma notation: n starts at 0, goes to infinity.

(z^n) times ((-1)^n) times ((n+2)/n)!, all divided by n!.

crap sorry this is wrong, look at scots. He has it right.
Then write out the sequence, find a gen. term.

Answer 4

What's a macclaurin series???

Answer 5

http://mathworld.wolfram.com/MaclaurinSeries.html

hope this helps. and don't worry; your world won't dissinigrate.