A) 1
B) n
C) \[\frac{1}{n}\]
D) \[\frac{2}{n+1}\]
Correct Answer: D
Solution :
Mean \[=\frac{1.1+\frac{1}{2}.2+\frac{1}{3}.3+\frac{1}{4}.4+\frac{1}{5}.5+.....+\frac{1}{n}.n}{1+2+3+.....+n}\] \[=\frac{1+1+1+1+.....+1}{\frac{n(n+1)}{2}}\]\[=\frac{n}{\frac{n(n+1)}{2}}=\frac{2}{n+1}\].
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