C0+C1+2.C2(3)+3.C3(32)+4.C4(33)+……+n.Cn3n-1=

Sum =C0+∑r=1n r Cr 3r-1 =1+n ∑r=1n Cr-1 3r-1 =1+n 1+3n-1 =1+n 4n-1

C0+C1+2.C2(3)+3.C3(32)+4.C4(33)+……+n.Cn3n-1=

Sum =C0+∑r=1n r Cr 3r-1 =1+n ∑r=1n Cr-1 3r-1 =1+n 1+3n-1 =1+n 4n-1

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$C_{0} + C_{1} + 2 . C_{2} (3) + 3 . C_{3} (3^{2}) + 4 . C_{4} (3^{3}) + \dots \dots + n . C_{n} 3^{n - 1} =$

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$n . 4^{n + 1} - 1$

$2 n . 4^{n - 1} + 1$

$n . 4^{n - 1} - 1$

$n . 4^{n - 1} + 1$

answer is A.

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Sum $= C_{0} + \sum_{r = 1}^{n} (r C_{r} 3^{r - 1})$

$= 1 + n \sum_{r = 1}^{n} C_{r - 1} 3^{r - 1} = 1 + n {(1 + 3)}^{n - 1} = 1 + n 4^{n - 1}$

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C0+C1+2.C2(3)+3.C3(32)+4.C4(33)+……+n.Cn3n-1=

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$C_{0} + C_{1} + 2 . C_{2} (3) + 3 . C_{3} (3^{2}) + 4 . C_{4} (3^{3}) + \dots \dots + n . C_{n} 3^{n - 1} =$