$$\sum_{ k = 1 }^{ n } k^2 = \overbrace{ 1^2 + 2^2 + \cdots + n^2 }^{ n } = \frac{ 1 }{ 6 } n ( n + 1 ) ( 2n + 1 )$$
$$ \int_0^\pi \sin{mx}\cos{nx} dx = \frac{1-(-1)^{m-n}}{m^2 - n^2} = \begin{cases} 0 & ( m-n \in even ) \\ \frac{2m}{m^2 - n^2} & ( m-n \in odd ) \end{cases} $$
$$ S_n = \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right) = 1 - \frac{1}{n+1} $$
$$ \begin{aligned} x_1 &= a_1 &+ b_1 &+ c_1 &+ d_1 \\ x_2 &= a_2 \\ x_3 &= &+ b_3 \\ x_4 &= a_4 & & &+ d_4 \end{aligned} $$
$$ \begin{aligned} \int \frac{1}{x(x+2)}dx &= \int \frac{1}{2}\left(\frac{1}{x}+\frac{-1}{x+2}\right) dx \\ &= \frac{1}{2}\int \frac{1}{x}dx - \frac{1}{2}\int \frac{1}{x+2}dx \\ &= \frac{1}{2}\left(\log |x| -\log|x+2|\right)+C \\ &= \frac{1}{2}\log \left|\frac{x}{x+2}\right|+C \end{aligned} $$
$$ \begin{aligned} 2x+6y+4(x+3y) &= 2(x+2y) \\ 2x+6y+4x+12y &= 2(x+2y) \\ 6x+18y &= 2(x+2y) \\ 6x+18y-18y &= 2x+4y-18y \\ 6x &= 2x-14y \\ 6x-2x &= 2x-2x-14y \\ 4x &= -14y \\ x &= \frac{-14y}{4} \end{aligned} $$
$$ \begin{aligned} x_t &= \phi_1 x_{t-1} &+ \phi_2 x_{t-2} &+ \phi_3 x_{t-3} &+ \epsilon_t \\ x_{t-1} &= 1 \cdot x_{t-1} \\ x_{t-2} &= &+1\cdot x_{t-2}\\ Y_t &= x_t & & &+ 0 \cdot \epsilon'_t \end{aligned} $$