= Solution
{c}
At $x=0$, the Fourier sum is $f(0)=1$, and only the odd cosine coefficients contribute. Therefore $1=1/4+(6/\pi^2)\sum_{k\geq0}(2k+1)^{-2}$, giving
$$
\boxed{\sum_{k=0}^\infty\frac1{(2k+1)^2}=\frac{\pi^2}{8}.}
$$
At $x=1/2$, the sum is $f(1/2)=1/2$. All nonzero cosine terms vanish, and the sine terms have alternating signs over odd indices. Thus $1/2=1/4+(1/\pi)\sum_{k\geq0}(-1)^k/(2k+1)$, so
$$
\boxed{\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac\pi4.}
$$
This is the <Leibniz formula for π> obtained from the same <Fourier series>.
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