A) An integer iff \[n\]is odd integer
B) An integer iff \[n\]is an even integer
C) Never integer
D) Always integer
Correct Answer: D
Solution :
| \[x+1={}^{n}{{C}_{n}}+{}^{n}{{C}_{n-1}}+{}^{n+1}{{C}_{n-1}}+....+{}^{2n-1}{{C}_{n-1}}\]\[={}^{2n}{{C}_{n}}\] |
| \[\therefore \,\,\frac{x+1}{n+1}=\frac{{}^{2n}{{C}_{n}}}{n+1}=\frac{{}^{2n+1}{{C}_{n+1}}}{2n+1}\in I\] |
| If \[2n+1\,\]& \[n+1\] are co-prime |
| Let \[2n+1=\lambda {{I}_{1}}\] & \[n+1=\lambda {{I}_{2}}\] |
| \[\therefore \,\,\,\frac{2n+1}{\lambda }-\frac{2\,\,(n+1)}{\lambda }={{I}_{1}}-2{{I}_{2}}\]\[-\frac{I}{\lambda }={{I}_{1}}-2{{I}_{2}}\in I\] |
| \[\therefore \,\,\,\lambda =\pm \,1\] |
| So \[2n+1\,\] & \[n+1\] are co-prime |
| \[\therefore \,\,\,\frac{x+1}{n+1}\] is always integer |
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