= Solution
For the <Fourier sine series> write $u_0(x)=\sum_{n\geq1}b_n\sin(n\pi x/L)$. <Orthogonality of trigonometric functions> gives
$$
b_n=\frac2L\int_0^L u_0(x)\sin\frac{n\pi x}L\,dx.
$$
In the second half of the integral put $z=L-x$; then $u_0(L-z)=\rho z$ and $\sin[n\pi(L-z)/L]=(-1)^{n+1}\sin(n\pi z/L)$. Hence
$$
b_n=\frac{2\rho}L[1+(-1)^{n+1}]\int_0^{L/2}x\sin\frac{n\pi x}L\,dx.
$$
Every even coefficient is zero. For odd $n$, part (a) or the supplied integral yields
$$
b_n=\frac{4L\rho}{n^2\pi^2}\sin\frac{n\pi}2.
$$
Putting $n=2m+1$ therefore gives
$$
\boxed{u_0(x)=\sum_{m=0}^{\infty}F_m\sin\frac{(2m+1)\pi x}L,\qquad
F_m=(-1)^m\frac{4L\rho}{(2m+1)^2\pi^2}.}
$$
This is the <Fourier sine series of a symmetric tent>. The absolute values of its terms are uniformly bounded by a summable sequence, giving <absolute convergence> and <uniform convergence>; the piecewise smooth odd extension and the <Dirichlet convergence theorem for Fourier series> identify its value with the continuous $u_0$, including the zero endpoints and the peak.
\b[The printed lower limit $m=1$ is an error.] The original PDF visibly has this lower limit. It omits the nonzero fundamental coefficient $F_0=4L\rho/\pi^2$. In particular, the printed series equals $u_0(x)-F_0\sin(\pi x/L)$, rather than $u_0(x)$. At $x=L/2$ it gives $L\rho/2-4L\rho/\pi^2$ instead of $L\rho/2$. The corrected lower limit is displayed explicitly above.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-main-tent-series.png]
{title=The tent Fourier series requires its fundamental mode}
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