Question

Va rog mult am mare nevoie de ajutor !!!!

Answer 1

1)
[tex]C_{n+1}^2= \frac{n(n+1)}{2} \\ 1-\frac{n(n+1)}{2}= \frac{(n-1)(n+2)}{n(n+1)} \\ x_n= \frac{n+2}{3n}\\ 1/2\leq \frac{n+2}{3n}\leq 2/3\\ 2\leq n\leq 4=\ \textgreater \ n\in\{2,3,4\}[/tex]
2)
[tex] \frac{C_{n+1}^{k}}{C_{n}^{k}} = \frac{n+1}{n+1-k} \\ P= \frac{2}{1} * \frac{4}{2}* \frac{6}{3} *...* \frac{2014}{1007} =2^{1007}[/tex]
3)
[tex](1+x)^{2013}=C_{2013}^0+C_{2013}^1x+C_{2013}^{2}x^2+...+C_{2013}^{2013}x^{2013}\\ Derivam\ egalitatea\ de\ mai\ sus:\\ 2013(1+x)^{2012}=C_{2013}^{1}+2xC_{2013}^{2}+3x^2C_{2013}^{3}+...+2013x^{2012}C_{2013}^{2013}\\ Pentru\ x=-1\\ 0=C_{2013}^{1}-2C_{2013}^{2}+3C_{2013}^{3}+...+2013C_{2013}^{2013}\\ S=0[/tex]