Find the Maclaurin Series for f(x) using the defination of a Maclaurin Series

f(x) = (1-x)^-2

f(x) = ln(1+x)

1) f(x) = (1 - x)^(-2)
f '(x) = 2(1 - x)^(-3)
f ''(x) = 3 * 2(1 - x)^(-4)
f '''(x) = 4 * 3 * 2(1 - x)^(-5)
...
f^(n)(x) = (n+1)! (1 - x)^(-n-2) for all n = 0, 1, 2, ...
==> f^(n)(0) = (n+1)!

So, (1 - x)^(-2) = Σ(n=0 to ∞) (n+1)! x^n/n! = Σ(n=0 to ∞) (n+1) x^n.
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2) f(x) = ln(1 + x) ==> f(0) = ln 1 = 0

f '(x) = 1/(1+x)
f ''(x) = -1/(1+x)^2
f '''(x) = 2/(1+x)^3
f ''''(x) = -3 * 2/(1+x)^4
...
f^(n)(x) = (-1)^(n+1) * (n-1)! / (1+x)^n
==> f^(n)(0) = (-1)^(n+1) * (n-1)! for all n = 1, 2, ...

So, ln(1 + x)
= 0 + Σ(n=1 to ∞) (-1)^(n+1) * (n-1)! x^n/n!
= Σ(n=1 to ∞) (-1)^(n+1) x^n/n.

I hope this helps!