A šta će ti opšti član?
Dati su ti, konkretno, drugi, treći i četvrti član, pa njih i pišemo:[dispmath]\begin{array}{l}
{n\choose 1}x^{n-1}y=240\\
{n\choose 2}x^{n-2}y^2=720\\
{n\choose 3}x^{n-3}y^3=1080
\end{array}[/dispmath]
[dispmath]\begin{array}{ll}
nx^{n-1}y=240 \\
\frac{n\left(n-1\right)}{2}x^{n-2}y^2=720\quad /\cdot 2 \\
\frac{n\left(n-1\right)\left(n-2\right)}{3!}x^{n-3}y^3=1080\quad /\cdot 6
\end{array}[/dispmath]
[dispmath]\begin{array}{ll}
nx^{n-1}y=240 & \left(1\right) \\
n\left(n-1\right)x^{n-2}y^2=1440 & \left(2\right) \\
n\left(n-1\right)\left(n-2\right)x^{n-3}y^3=6480 & \left(3\right)
\end{array}[/dispmath]
[dispmath]\frac{\left(2\right)}{\left(1\right)}\quad\Rightarrow\quad\frac{\cancel n\left(n-1\right)x^{n-2}y^2}{\cancel nx^{n-1}y}=\frac{1440}{240}=6[/dispmath][dispmath]\Rightarrow\quad\left(n-1\right)\frac{y}{x}=6\quad\Rightarrow\quad\frac{y}{x}=\frac{6}{n-1}\quad\left(4\right)[/dispmath]
[dispmath]\frac{\left(3\right)}{\left(2\right)}\quad\Rightarrow\quad\frac{\cancel n\cancel{\left(n-1\right)}\left(n-2\right)x^{n-3}y^3}{\cancel n\cancel{\left(n-1\right)}x^{n-2}y^2}=\frac{6480}{1440}=\frac{9}{2}[/dispmath][dispmath]\Rightarrow\quad\left(n-2\right)\frac{y}{x}=\frac{9}{2}\quad\Rightarrow\quad\frac{y}{x}=\frac{9}{2\left(n-2\right)}\quad\left(5\right)[/dispmath]
[dispmath]\left(4\right),\left(5\right)\quad\Rightarrow\quad\frac{6}{n-1}=\frac{9}{2\left(n-2\right)}[/dispmath][dispmath]\cdots[/dispmath][dispmath]\underline{n=5}\quad\left(6\right)[/dispmath]Sada [inlmath]\left(6\right)[/inlmath] uvrstimo u [inlmath]\left(4\right)[/inlmath] ili u [inlmath]\left(5\right)[/inlmath] i dobijemo da je [inlmath]\frac{y}{x}=\frac{3}{2}[/inlmath], a zatim [inlmath]\left(6\right)[/inlmath] uvrstimo u [inlmath]\left(1\right)[/inlmath], nakon čega imamo sistem od dve jednačine s dve nepoznate, [inlmath]x[/inlmath] i [inlmath]y[/inlmath]... Njegovim rešavanjem se dobije [inlmath]x=2[/inlmath], [inlmath]y=3[/inlmath].