You need an eagle eye for this

$\large y = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \ldots$ Above series is expansion of $\cos(2x)$
$\large y = \cos(2x)$ $\large \dfrac{dy}{dx} = -2\sin(2x)$ $\large \dfrac{d^2y}{dx^2} = -4\cos(2x) = -4\cos(2\pi) = -4$

Simply the function of 'x' mentioned is the expansion of cos2x , hence double derivative of cos2x at x=pi is [-4].

The expansion of the cosine function around $x = 0$ is $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -+ \cdots + (-)^n\frac{x^{2n}}{(2n)!} + \cdots.$ Thus the given series is equal to $y = \cos(2x).$ The derivative is therefore $\frac {d^2}{dx^2}\cos(2x) = \frac d{dx}(-2\sin 2x) = -4\cos 2x,$ and substituting $x = \pi$ gives $-4\cos 2\pi = \boxed{-4}.$

Alternative:

The second derivative of the term $T = \frac{(2x)^n}{n!} = \frac{2^n}{n!}x^n$ is $\frac{d^2T}{dx^2} = \frac{2^n}{n!}\cdot n\cdot (n-1)\cdot x^{n-2} = \frac{2^n}{(n-2)!}x^{n-2} = 4\cdot\frac{(2x)^{n-2}}{(n-2)!},$ which is equal to the previous term except for the sign. Therefore $\frac{d^2y}{dx^2} = -4y$ for every value of $x$ .

Of course, to see that $y = 1$ when $x = \pi$ , we still need to recognize that this series describes the cosine function.