A) \[2r+3{{r}^{2}}\]
B) \[3{{r}^{2}}-4r+1\]
C) \[6r-1\]
D) \[4r+1\]
Correct Answer: C
Solution :
Let \[{{S}_{n}}=2n+3{{n}^{2}}\] \[\therefore \] \[{{S}_{n-1}}=2(n-1)+3{{(n-1)}^{2}}\] \[=2n-2+3{{n}^{2}}+3-6n\] \[=3{{n}^{2}}-4n+1\] \[\therefore \] \[{{T}_{n}}={{S}_{n}}-{{S}_{n-1}}\] \[=2n+3{{n}^{2}}-(3{{n}^{2}}-4n+1)=6n-1\] \[\therefore \] \[{{T}_{r}}=6r-1\]
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