• # question_answer If $x={}^{n}{{C}_{n\,-1}}+{}^{n+1}{{C}_{n\,-1}}+.....+{}^{2n\,-1}{{C}_{n\,-1}}$ then $\frac{x+1}{n+1}$ is A) an integer iff n is odd integer B) an integer iff n is an even integer C) never integer D) always integer

 $x+1={}^{n}{{C}_{n}}+{}^{n}{{C}_{n\,-1}}+{}^{n\,+\,1}{{C}_{n\,-1}}+.......+{}^{2n\,-1}{{C}_{n\,-1}}={}^{2n}{{C}_{n}}$ $\therefore$$\frac{x+1}{n+1}=\frac{^{2n}{{C}_{n}}}{n+1}=\frac{^{2n\,\,+\,1}{{C}_{n\,+\,1}}}{2n+1}\in I$ if $2n+1$ & $n+1$ are co-prime Let        $2n+1=\lambda \,{{I}_{1}}$ & $n+1=\lambda \,{{I}_{2}}$ $\therefore$$\frac{2n+1}{\lambda }-\frac{2\,(n+1)}{\lambda }={{I}_{1}}-2{{I}_{2}}$
 $-\frac{1}{\lambda }={{I}_{1}}-2{{I}_{2}}\in I$ $\therefore$      $\lambda =\pm \,1$ So $2n+1$ & $n+1$ are co-prime $\therefore$      $\frac{x+1}{n+1}$ is always integer