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 \[2n+1=\lambda \,{{I}_{1}}\] & \[n+1=\lambda \,{{I}_{2}}\] |
| \[\therefore \]\[\frac{2n+1}{\lambda }-\frac{2\,(n+1)}{\lambda }={{I}_{1}}-2{{I}_{2}}\] |
| \[-\frac{1}{\lambda }={{I}_{1}}-2{{I}_{2}}\in I\] |
| \[\therefore \] \[\lambda =\pm \,1\] |
| So \[2n+1\] & \[n+1\] are co-prime |
| \[\therefore \] \[\frac{x+1}{n+1}\] is always integer |
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