Question 1195861
<pre>

In your notes somewhere, you have shown that the Taylor series for ln(1-x) is

{{{ln(1-x)}}}{{{""=""}}}{{{-sum(x^n/n,n=1,infinity)}}}

Maybe your professor expected you to have memorized it, or to have access to it.
Surely he or she didn't expect you to find the Taylor series for every one of
the choices.  So I'll assume you have either memorized the above or have access
to it.

Substitute x<sup>2</sup> for x

{{{ln(1-x^2)}}}{{{""=""}}}{{{-sum( (x^2)^n/n,n=1,infinity)}}}

{{{ln(1-x^2)}}}{{{""=""}}}{{{-sum(x^(2n)/n,n=1,infinity)}}}

Just multiply both sides by -1

{{{-ln(1-x^2)}}}{{{""=""}}}{{{sum(x^(2n)/n,n=1,infinity)}}}

So it's D.

Edwin </pre>