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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ them the value of $C_{0} + 2 C_{1} + 3 C_{2} + \dots + (n + 1) C_{n}$ will be

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(n+2)2n−1

(n+1)2n

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(n+2)2n

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detailed solution

Correct option is A

Since, x(1+x)n=xC0+C1x2+C2x3+…+Cnxn+1 On differentiating w.r.t. x, we get (1+x)n+nx(1+x)n−1=C0+2C1x+3C2x2+…+(n+1)CnxnPut x = L, we get C0+2C1+3C2+…+(n+1)Cn=2n+n2n−1=2n−1(n+2)

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If(1+x)n=C0+C1x+C2x2+…+Cnxn them the value of C0+2C1+3C2+…+(n+1)Cn will be

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ them the value of $C_{0} + 2 C_{1} + 3 C_{2} + \dots + (n + 1) C_{n}$ will be