A) \[n\]
B) \[2n-1\]
C) \[{{n}^{2}}\]
D) \[{{n}^{3}}\]
Correct Answer: C
Solution :
[c] \[{{1}^{st}}\]term =1 =2-1, \[{{2}^{nd}}\]term =3 =4-1, \[{{n}^{th}}\]term is 2n-1 Therefore, \[\sum{(2n-1)=2.\frac{n(n+1)}{2}}-n={{n}^{2}}+n-n={{n}^{2}}\]
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