SOLUTION: This is a calculus question. Could you explain question 18 on https://www.math.purdue.edu/php-scripts/courses/oldexams/serve_file.php?file=16600FE-F2016.pdf ? I know the answer is

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



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 x2 for x





Just multiply both sides by -1



So it's D.

Edwin 

Every/any/each Calculus student knows (or must know - this knowledge is a pre-requisite) that

    ln(1+y) =  -  +  -  + . . .     (1)

for all real values y, |y| < 1.  It is called  " the Maclaurin series of  f(y) = ln(1+y) ".



Replace (substitute) here y = -x^2.  You will get

    ln(1-x^2) =  -  -  -  - . . .     (2)



Right side is your series with the sign "-, minus".



So, multiply both sides of identity (2) by (-1), and you will get the identity (D), which you need

     =  +  +  +  + . . .     (3)