A) 2
B) 3
C) 1
D) Any natural number
Correct Answer: D
Solution :
(d): Mean \[=\frac{1+3+5+7+......(2n-1)}{n}=n\] Or, \[[1+2+3+.....(2n-1)]\]\[-[2+4+6+....(2n-2)]={{n}^{2}}\] Or, \[\frac{(2n-1)(2n)}{2}-2[1+2+....(n-1)]={{n}^{2}}\] Or, \[n(2n-1)-\frac{2\times (n-1)n}{2}={{n}^{2}}\] Or, \[{{n}^{2}}={{n}^{2}}\Rightarrow \] for all natural numbers ?n? is true.
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