If 1.3 + 2.32 + … + n.3n \(= \frac{{\left( {2n - 1} \right){3^a} + b}}{4}\) then a and b are respectively

If 1.3 + 2.32 + … + n.3n \(= \frac{{\left( {2n - 1} \right){3^a} + b

1.	If 1.3 + 2.32 + … + n.3n \(= \frac{{\left( {2n - 1} \right){3^a} + b}}{4}\) then a and b are respectively
A.	n, 2
B.	n, 3
C.	n + 1, 2
D.	n + 1, 3
Answer» E.

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If 1.3 + 2.32 + … + n.3n \(= \frac{{\left( {2n - 1} \right){3^a} + b}}{4}\) then a and b are respectively

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