C1+2⋅C2+3⋅C3+…..+n⋅Cn=

⇒ddxC0+C1x+C2x2+…+Cnxn=ddx(1+x)n⇒C1+2⋅C2x+3⋅C3x2+….+n⋅Cnxn−1=n⋅(1+x)n−1x=1⇒C1+2⋅C2+3⋅C3+….+n⋅Cn=n⋅2n−1

C1+2⋅C2+3⋅C3+…..+n⋅Cn=

⇒ddxC0+C1x+C2x2+…+Cnxn=ddx(1+x)n⇒C1+2⋅C2x+3⋅C3x2+….+n⋅Cnxn−1=n⋅(1+x)n−1x=1⇒C1+2⋅C2+3⋅C3+….+n⋅Cn=n⋅2n−1

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$C_{1} + 2 \cdot C_{2} + 3 \cdot C_{3} + \dots . . + n \cdot C_{n} =$

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$n \cdot 2^{n}$

$n . 2^{n - 1}$

$(n + 1) 2^{n}$

$(n - 1) \cdot 2^{n}$

answer is B.

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Detailed Solution

$\begin{array}{l} \Rightarrow \frac{d}{dx} \{C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}\} = \frac{d}{dx} \{(1 + x)^{n}\} \\ \Rightarrow C_{1} + 2 \cdot C_{2} x + 3 \cdot C_{3} x^{2} + \dots . + n \cdot C_{n} x^{n - 1} = n \cdot (1 + x)^{n - 1} \\ x = 1 \Rightarrow C_{1} + 2 \cdot C_{2} + 3 \cdot C_{3} + \dots . + n \cdot C_{n} = n \cdot 2^{n - 1} \end{array}$