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\begin{document}
 \section{Exercise 1.4}
 \begin{math}
  \begin{cases}
   x_1 + x_2 = 1 \\
   x_2 + x_3 = 3 \\
   x_3 + x_4 = 5 \\
   x_{n - 1} + x_n = 2n - 3 \\
   2x_1 + x_n = 2n - 1
  \end{cases}
 \end{math}
 \subsection{a)}
 In case when n is even, the last equation is odd numbered. The right sides of even and odd numbered equations both form arithmetic series, whose sums can be calculated. \\
 The sum of all oddly numbered right sides: \\
 \begin{math}
  a_1 = 1 \\
  a_n = 2n - 3 \\
  n_n = \frac{n}{2} \text{ - count of odd numbered equations} \\
  S_n = \frac{1 + 2n - 3}{2} \cdot \frac{n}{2} = \frac{2n^2 - 2n}{4} = \frac{n^2 - n}{2}
 \end{math} \\
 Sum of all even numbered right sides: \\
 \begin{math}
  a_1 = 3 \\
  a_n = 2n - 1 \\
  n_p = \frac{n}{2} \text{ - count of odd numbered equations} \\
  S_n = \frac{3 + 2n - 1}{2} \cdot \frac{n}{2} = \frac{2n^2 + 2n}{4} = \frac{n^2 + n}{2}
 \end{math} \\
 By summing all even and odd numbered equations and leaving the last equation we can write down a simplified equation system, where the first line is the sum of even numbered equations, second line is the sum of all odd numbered ones and the last line is the last line from the original system.\\
 \begin{math}
  \begin{cases}
   x_1 + x_2 + x_3 + x_4 + ... + x_{n - 1} + x_n &= \frac{n^2 - n}{2} \\
   2x_1 + x_2 + x_3 + x_4 + ... + x_{n - 1} + x_n &= \frac{n^2 + n}{2} \\
   2x_1 + x_n &= 2x - 1
  \end{cases} \\
 \end{math}
 By subtracting the second line from the first we get: \\
 \begin{math}
  \begin{cases}
   -x_1 = -n \\
   2x_1 + x_n = 2n - 1 \\
  \end{cases} \\
  x_n = -1 \\
  2x_1 - 1 = 2n - 1 \\
  x_1 = n
 \end{math}
 \subsection{b)}
 For odd n, the second to last equation is even numbered. We add all odd numbered equations to the first one: \\
 \begin{math}
  x_1 + x_2 + x_3 + x_4 + ... + x_{n - 2} + x_{n - 1} = \frac{n(2n - 3)}{2} + 1
 \end{math}
 \\
 We add together all even numbered equations as well: \\
 \begin{math}
  x_2 + x_3 + x_4 + ... + x_{n - 1} + x_n = \frac{n(2n - 3)}{2}
 \end{math} \\
 And subtract it from the first equation: \\
 \begin{math}
  x_1 + x_2 + x_3 + x_4 + ... + x_{n - 2} + x_{n - 1} - (x_2 + x_3 + x_4 + ... + x_{n - 1} + x_n) = \frac{n(2n - 3)}{2} + 1 - \frac{n(2n - 3)}{2} \\
  x_1 - x_n = 1
 \end{math} \\
 We create an equation system from the resulting equation and the last from the original system and then we solve it: \\
 \begin{math}
  \begin{cases}
   x_1 - x_n = 1 \\
   2x_1 + x_n = 2n - 1
  \end{cases} \\
  2x_1 + x_1 + x_n - x_n = 2n - 1 + 1 \\
  3x_1 = 2n \\
  x_1 = \frac{2n}{3} \\
  \frac{4n}{3} + x_n = 2n - 1 \\
  x_n = \frac{2n}{3} - 1
 \end{math}
\end{document}