A) \[n\,(2n+1)\]
B) \[2n\,(2n-1)\]
C) \[(2n+1)(2n-1)\]
D) \[4{{n}^{2}}+1\]
Correct Answer: C
Solution :
Since, given series is \[(1\times 3)+(3\times 5)+(5\times 7)+(7\times 9)+.......+(2n-1)(2n+1)\] Hence \[{{n}^{th}}\] term of the series is \[(2n+1)(2n-1)\].
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