A) \[{{n}^{2}}-1\]
B) \[{{(2n-1)}^{2}}\]
C) \[{{n}^{2}}\]
D) \[{{n}^{2}}+1\]
Correct Answer: C
Solution :
Given that \[{{T}_{n}}=2n-1\] First term \[a=2.1-1=1\] Second term \[b=2\ .\ 2-1=3\] Third term \[c=2\ .\ 3-1=5\] Therefore sequence is \[1,\ 3,\ 5,.......2n-1\]. Now sum of the first \[n\] terms is \[{{S}_{n}}=\frac{n}{2}[a+l]\] \[=\frac{n}{2}[1+2n-1]=\frac{n}{2}\ .\ 2n={{n}^{2}}\] Aliter: Since \[{{T}_{n}}=2n-1\] \[\Rightarrow {{S}_{n}}=\Sigma {{T}_{n}}=2\Sigma n-\Sigma \ 1=n(n+1)-n={{n}^{2}}\]
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