a hint for this Taylor series$ \frac{\cos\left(2x\right)-1}{x^2}$ Compute the first three terms (nonzero)  $\frac{\cos\left(2x\right)-1}{x^2}$
the first term is $\cos \left(2\right)-1$ but in the answer, the first term that I have to choose is... $-2$ or $2$ or $-1/2$ 
UPDATE: Use the substitution method​     
 A: Hint
$${d^n\over dx^n}\cos x=\begin{cases}
-\sin x&,\quad n=4k+1\\
-\cos x&,\quad n=4k+2\\
\sin x&,\quad n=4k+3\\
\cos x&,\quad n=4k+4\\
\end{cases}$$therefore
$${d^n\over dx^n}\cos x\Bigg|_{x=0}=\begin{cases}
0&,\quad n=4k+1\\
-1&,\quad n=4k+2\\
0&,\quad n=4k+3\\
1&,\quad n=4k+4\\
\end{cases}$$
Also$$
f( x)=\sum_{n=0}^{\infty}{f^{(n)}(0)\over n!}x^n
$$
A: Hint: 
If the constant term is chosen between $-2$ or $2$ or $-1/2$, most likely you are asked to compute Taylor series about $x = 0$, rather than $x = 1$ that gives $\cos(2) - 1$.
Use L'Hôpital's rule to verify the constant term.
A: Assuming that you want to find the Maclaurin series which is the Taylor series expansion about $0$, you can start from
$$\cos(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+ \dots$$
therefore
\begin{align}\cos(2x)=\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^{2n}}{(2n)!}&=1-\frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!}+ \dots\\&=1-2x^2+\frac{2x^4}{3}-\frac{4x^6}{45}+\dots\end{align}
and
\begin{align}\frac{\cos(x)}{x^2}=\sum_{n=0}^{\infty}(-1)^n\frac{2^{2n}(x)^{2n-2}}{(2n)!}&=\frac{1}{x^2}-\frac{(2x)^2}{2!(x^2)}+\frac{(2x)^4}{4!(x^2)}-\frac{(2x)^6}{6!(x^2)}+ \dots \\&=\frac{1}{x^2}-\frac{2^2}{2!}+\frac{2^4 x^2}{4!}-\frac{2^6 x^4}{6!}+ \dots
\\&=\frac{1}{x^2}-2+\frac{2x^2}{3}-\frac{4x^4}{45}+ \dots\end{align}
will allow you to analyze terms of the Taylor series for $\frac{cos(2x)-1}{x^2}$.
A: Make the substitution
$$y=2x$$
Then you can use the standard MacLaurin series
