,

\begin{eqnarray*}
2012 & = & \frac{a_{1}+a_{3}}{a_{2}+a_{4}}+\frac{a_{3}+a_{5}}{a_{4}+a_{6}}+\frac{a_{5}+a_{7}}{a_{6}+a_{8}}+\cdots+\frac{a_{2011}+a_{2013}}{a_{2012}+a_{2014}}\\
2012 & = & \frac{a_{1}\left(1+r^2\right)}{a_{1}r\left(1+r^2\right)}+\frac{a_{1}r^{2}\left(1+r^2\right)}{a_{1}r^{3}\left(1+r^2\right)}+\frac{a_{1}r^{4}\left(1+r^2\right)}{a_{1}r^{5}\left(1+r^2\right)}+\cdots+\frac{a_{1}r^{2010}\left(1+r^2\right)}{a_{1}r^{2011}\left(1+r^2\right)}\\
2012 & = & \frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\cdots+\frac{1}{r}\\
2012 & = & \frac{1006}{r}\\
r & = & \frac{1}{2}
\end{eqnarray*}

,

กำหนดให้ 

$$S=1+4r+10r^{2}+19r^{3}+\cdots\quad\cdots(1)$$

คูณ $r$ ตลอดสมการที่ $(1)$

\begin{align}
S &= &1+  &4r + 10r^2 + 19r^3+ \cdots\quad\cdots(2)\\
rS &=&&r  +  4r^2 + 10r^3 +19r^4+\cdots\quad\cdots(3)\\
\end{align}

$(2)-(3):$

\begin{eqnarray*}
S-rS & = & 1+3r+6r^{2}+9r^{3}+\cdots\\
S\left(1-r\right) & = & 1+3r+6r^{2}+9r^{3}+\cdots
\end{eqnarray*}

ให้ 

$$T=1+3r+6r^{2}+9r^{3}+\cdots$$

คูณตลอดด้วย $r$ 

\begin{align}
T & = & 1+&3r+6r^{2}+9r^{3}+\cdots\quad\cdots\left(4\right)\\
rT & = && r+3r^{2}+6r^{3}+9r^{4}+\cdots\quad\cdots\left(5\right)
\end{align}

$(4)-(5):$

\begin{eqnarray*}
T-rT & = & 1+2r+3\left(r^{2}+r^{3}+r^{4}+\cdots\right)\\
T\left(1-r\right) & = & 1+2r+3\left(\frac{r^{2}}{1-r}\right)\\
T & = & \frac{1+2r+3\left(\frac{r^{2}}{1-r}\right)}{1-r}
\end{eqnarray*}

แทนค่า $r=\frac12$ ลงในสมการ $T$

\begin{eqnarray*}
T & = & \frac{1+2\left(\frac{1}{2}\right)+3\left(\frac{\left(\frac{1}{2}\right)^{2}}{1-\frac{1}{2}}\right)}{1-\frac{1}{2}}\\
 & = & 7
\end{eqnarray*}

แทนค่า $r=\frac12$ และ $T=7$ ลงในสมการ $S$

\begin{eqnarray*}
S\left(1-r\right) & = & 1+3r+6r^{2}+9r^{3}+\cdots\\
S\left(1-r\right) & = & T\\
S & = & \frac{T}{1-r}\\
 & = & \frac{7}{1-\frac{1}{2}}\\
 & = & 14
\end{eqnarray*}