The sum of series $1\cdot 3+2\cdot 3^2+3\cdot 3^3+\dots +\mathrm{n}\cdot 3^{\mathrm{n}}$ is

detailed solution

Correct option is B

Let the given statement be P(n). P(n):1⋅3+2⋅32+3⋅33+…+n⋅3n=(2n−1)3n+1+34Step l: Let it is true for n = 1,      P(1):(2⋅1−1)31+1+34=32+34=9+34=124    ⇒3=1.3  which is trueStep II: For n=k,  i.e. 1⋅3+2⋅32+3⋅33+…+k⋅3k=(2k−1)3k+1+34-----iStep III: For n=k+1,1⋅3+2⋅32+3⋅33+…+k⋅3k+(k+1)3k+1 =(2k−1)3k+1+34+(k+1)3k+1[ using Eq. (i)]  =(2k−1)3k+1+3+4(k+1)3k+14On taking 3k+1 common in first and last term of numerator part, =3k+1(2k−1+4k+4)+34=3k+1(6k+3)+34On taking 3 common in first term of numerator part,=3k+1⋅3(2k+1)+34=3(k+1)+1(2k+2−1)+34={2(k+1)−1}3(k+1)+1+34Therefore, P(k + 1) is true when P(k) is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers n.