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${Series}^{n} C_{1} + 1 \cdot^{n} C_{2} + 2 \cdot^{n} C_{3} + \dots + n \cdot^{n} C_{n}$ is equal to

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n2n−1

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Correct option is A

Since, (1+x)n=nC0+nC1x+nCx2+…+nCnxn Ondifferentiatingw.r.t., x, weget n(1+x)n−1=0+nC1+2⋅x⋅nC2+…+n⋅nCnxn−1Put x=1, we get n(1+1)n−1=nC1+2⋅nC2+…+n⋅nCn⇒ n2n−1=nC1+2⋅nC2+…+n⋅nCn

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Series nC1+1⋅nC2+2⋅nC3+…+n⋅nCn is equal to

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${Series}^{n} C_{1} + 1 \cdot^{n} C_{2} + 2 \cdot^{n} C_{3} + \dots + n \cdot^{n} C_{n}$ is equal to