A) n
B) 1/n
C) \[\frac{1}{n+1}\]
D) \[\frac{1}{n-1}\]
Correct Answer: C
Solution :
Trick: Put \[n=\] 1, 2 At \[n=1,{{\,}^{1}}{{C}_{0}}-\frac{1}{2}{{\,}^{1}}{{C}_{1}}=1-\frac{1}{2}=\frac{1}{2}\] At \[n=2,\,{{\,}^{2}}{{C}_{0}}-\frac{1}{2}\,{{\,}^{2}}{{C}_{1}}+\frac{1}{3}{{\,}^{2}}{{C}_{2}}=1-1+\frac{1}{3}=\frac{1}{3}\] which is given by option C.
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