We know the orthogonality condition
$$\int_0^1\sin(n\pi x)\sin(m\pi x)dx = \begin{cases} 0 & \text{ if } n \neq m\\ \frac12 & \text{ if } n = m\\ \end{cases} $$
From earlier in the text we had
$$\sum_{n=1}^\infty A_n \int_0^1\sin(n\pi x)\sin(m\pi x) dx = \int_0^1 g(x)\sin(m\pi x) dx$$
which by the first line gives
$$A_n = 2\int_0^1g(x)\sin(n\pi x) dx.$$
The summation symbol disappears from in front of $A_n$ when arriving at the last line. I am hoping to learn why. Thank you in advance for your time.
Your first line is the same as
$$\int_0^1\sin(n\pi x)\sin(m\pi x)dx = \frac12 \delta_{nm},$$
where $\delta$ is the Kronecker Delta. Now your second line is
$$\sum_{n=1}^\infty A_n \int_0^1\sin(n\pi x)\sin(m\pi x) dx = \int_0^1 g(x)\sin(m\pi x) dx$$
and becomes
$$\sum_{n=1}^\infty A_n \frac12 \delta_{nm} = \int_0^1 g(x)\sin(m\pi x) dx$$
Since only the $n=m$ term on the left hand side is nonzero (by the definition of $\delta$), this is
$$A_m \frac12 = \int_0^1 g(x)\sin(m\pi x) dx.$$
Note that the solution was presented as
$$A_n = 2\int_0^1g(x)\sin(n\pi x) dx,$$
which has the index $m$ switched to $n$, causing some confusion for beginners. But all of our statements are true for all integer $n,m$, so changing the name of the index doesn't matter.