Science:Math Exam Resources/Courses/MATH103/April 2015/Question 01 (e) (iii)/Solution 2

Again, we can find the answer by eliminating the options. The remaining options are $\cos x,\ln(1+x)$ and ${\frac {1}{(1-x)^{2}}}$.

At $x=0$, $\ln(1+x)$ is equal to 0, however the given series at $x=0$ is equal to 1.

$\cos x$ although is equal to 1 at $x=0$, its Taylor series does not have any odd degree term because its 1st, 3rd, 5th, ... derivative vanish at $x=0$, so the only option is ${\frac {1}{(1-x)^{2}}}$.