Question

Ex 3 va rog sa ma ajutați

Answer 1

 

$\displaystyle\\a)\\\\C_n^{n-2}+2n=9\\\\C_n^2+2n=9\\\\\frac{A_n^2}{P_2}+2n=9\\\\\frac{n(n-1)}{1\times2}+2n=9\\\\\frac{n^2-n}{2}+2n=9~~~\Big|\times2\\\\n^2-n+4n=18\\\\n^2+3n-18=0\\\\n_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-3\pm\sqrt{9+4\times18}}{2}=\\\\ =\frac{-3\pm\sqrt{9+72}}{2}=\frac{-3\pm\sqrt{81}}{2}=\frac{-3\pm9}{2}\\\\n_1=\frac{-3+9}{2}=\frac{6}{2}=\boxed{\bf3}\\\\n_2=\frac{-3-9}{2}=-6\notin R \\\\\implies \boxed{\bf n=3}~~~\text{Solutie unica}$

$\displaystyle\\b)\\\\3C_{n+1}^2-2A_n^2=n\\\\3\times\frac{A_{n+1}^2}{1\times2}-2A_n^2=n\\\\3\times\frac{(n+1)\times n}{1\times2}-2\times n(n-1)=n\\\\3\times\frac{(n+1)\times n}{2}-2\times n(n-1)=n~~~\Big|\times2\\\\3(n+1)\times n-4\times n(n-1)=2n\\\\3n^2+3n-4n^2+4n-2n=0\\\\-n^2+5n=0~~~\Big|\times(-1)\\n^2-5n=0\\\\n(n-5)=0\\n=0 \text{nu este admis}\\\\n-5=0\\\\\boxed{\bf n=5}$