A) 1
B) \[n-\frac{1}{n}\]
C) \[1-\frac{1}{{{n}^{2}}}\]
D) \[n-\frac{1}{{{n}^{2}}}\]
Correct Answer: C
Solution :
[c] Sum of the n numbers \[=\left( 1-\frac{1}{n} \right)+(1+1+...........+\overline{n-1}\,\,\,times)\] \[=\frac{n-1}{n}+(n-1)=(n-1)\left( 1+\frac{1}{n} \right)=\frac{{{n}^{2}}-1}{n}\] Mean (Average) \[=\frac{{{n}^{2}}-1}{n}.\frac{1}{n}=\frac{{{n}^{2}}-1}{{{n}^{2}}}\] \[=1-\frac{1}{{{n}^{2}}}\]You need to login to perform this action.
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