Maclaurin series?! Calc!

Please find the Maclaurin series for the following functions'

3/(1-2x)

1/(1+x)^3

Thank you!

Maclaurin Series for f(x)
f(x) = Sum from n = 0 to INF {f^n (0) (x^n) / n!} where f^n (0) is the n-th derivative of f at 0/

1) Let f(x) = 3/(1 - 2x). Compute some derivatives until we find a pattern.

f ' (x) = [(1-2x)(0) - (3)(-2)] / (1-2x)^2 = 6/(1 - 2x)^2

f '' (x) = [(1-2x)^2 (0) - 6(2)(1-2x)(-2)] / (1-2x)^4
f '' (x) = [24(1 - 2x)] / [1 - 2x]^4 = 24 / (1 - 2x)^3

f ''' (x) = [(1-2x)^3 (0) - 24(3)(1-2x)^2 (-2)] / [1-2x]^6
f ''' (x) = [144 (1-2x)^2] / [1 - 2x]^6 = 144 / (1 - 2x)^4

We see a pattern here:

f^n (x) = [3 * 2^(n) * n!] / (1 - 2x)^n

Now let x = 0.

f^n (0) = 3 * 2^n * n!

Plug into our Maclaurin Sum

f(x) = Sum from n = 0 to INF {3 * 2^n * n! * x^n / n!}
f(x) = Sum from n = 0 to INF {3 * 2^n * x^n}

2) Let g(x) = 1/(1 + x)^3 = (1 + x)^-3

g ' (x) = -3(1 + x)^(-4)
g '' (x) = 12(1 + x)^(-5)
g ''' (x) = -60(1 + x)^(-6)

g^n (x) = (-1)^n (1/2)(n + 2)! (1 + x)^(-3 - n)
g^n (0) = (-1)^n (1/2)(n + 2)!

g(x) = Sum from n = 0 to INF {(-1)^n * (1/2) * (n+2)! * x^n / n!}

Note that (n+2)! / n! = (n+2)(n+1)n! / n! = (n+2)(n+1)

g(x) = Sum from n = 0 to INF {(-1)^n * (1/2) * (n+2)(n+1) * x^n}

Use the geometric series 1/(1 - t) = Σ(n = 0 to ∞) t^n.

1) Let t = 2x:
1/(1 - 2x) = Σ(n = 0 to ∞) 2^n x^n

Multiply both sides by 3:
3/(1 - 2x) = Σ(n = 0 to ∞) 3 * 2^n x^n.
----------------------------
2) Differentiate both sides of the geometric series:
1/(1 - t)^2 = Σ(n = 1 to ∞) nt^(n-1)

Differentiate both sides again:
2/(1 - t)^3 = Σ(n = 2 to ∞) n(n-1)t^(n-2)

Let t = -x:
2/(1 + x)^3 = Σ(n = 2 to ∞) n(n-1)(-1)^(n-2) x^(n-2)
................= Σ(n = 2 to ∞) (-1)^n * n(n-1) x^(n-2), since (-1)^2 = 1

Divide both sides by 2:
1/(1 + x)^3 = Σ(n = 2 to ∞) (-1)^n * (1/2)n(n-1) x^(n-2).
-------------------------
I hope this helps!

The expansion of (1+a)^n, when -1 1)=3(1-2x)^-1, put a=-2x, n=-1 in the above to get 3(1+2x+4x^2+....2^rx^r+...)
Multiply each term if you like.
2) Put n=-3 and a=x