= Solution
For the real even <function>, let $s_N=a_0/2+\sum_{n=1}^Na_n\cos nx$. <Fourier orthogonality> gives $\int_{-\pi}^{\pi}\cos nx\,dx=0$ for $n\geq1$ and $\int_{-\pi}^{\pi}\cos nx\cos mx\,dx=\pi\delta_{nm}$. Consequently
$$
\frac1\pi\int_{-\pi}^{\pi}s_N^2\,dx=\frac{a_0^2}{2}+\sum_{n=1}^Na_n^2.
$$
For an even <square-integrable function>, the complete <Fourier cosine series> converges in the <L2 norm>; hence its norms converge. Taking $N\to\infty$ proves <Parseval's identity> in the required normalization:
$$
\boxed{\frac1\pi\int_{-\pi}^{\pi}f(x)^2\,dx=\frac{a_0^2}{2}+\sum_{n=1}^{\infty}a_n^2}.
$$
This limit argument avoids multiplying an infinite series without a justification of <L2 norm> convergence.
For the quadratic <function>, the <Fourier coefficients> are
$$
a_0=\frac1\pi\int_{-\pi}^{\pi}x^2\,dx=\frac{2\pi^2}{3},\qquad
a_n=\frac2\pi\int_0^\pi x^2\cos nx\,dx=\frac{4(-1)^n}{n^2}.
$$
The last equality follows from two <integrations by parts>, using $\sin n\pi=0$. Thus
$$
\boxed{x^2=\frac{\pi^2}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx\quad(-\pi\leq x\leq\pi)}.
$$
This <Fourier series> is <absolutely convergent>; the periodically extended quadratic is continuous, so it represents its values also at the endpoints. Substituting its coefficients into <Parseval's identity> gives
$$
\frac{2\pi^4}{5}=\frac{2\pi^4}{9}+16\sum_{n=1}^{\infty}\frac1{n^4},
\qquad \boxed{\sum_{n=1}^{\infty}\frac1{n^4}=\frac{\pi^4}{90}}.
$$
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