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

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Question 1195861: 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 D but don't know how to get there. Please attach an image of your work as a reply.
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

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 

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
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 D but don't know how to get there.
Please attach an image of your work as a reply.
~~~~~~~~~~~~~~~~~~~

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)

Solved and explained.



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