If (1+x)n=C0+C1x+C2x2+…+Cnxn, then C0+3C1+5C2+…+(2n+1)Cn=

E=Cn+C1+C2+…+Cn+2C1+2C2+…nCn=2n+2·n2n−1=(n+1)2n

If (1+x)n=C0+C1x+C2x2+…+Cnxn, then C0+3C1+5C2+…+(2n+1)Cn=

E=Cn+C1+C2+…+Cn+2C1+2C2+…nCn=2n+2·n2n−1=(n+1)2n

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} + 3 C_{1} + 5 C_{2} + \dots + (2 n + 1) C_{n} =$

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

$n . 2^{n}$

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

n+1

answer is C.

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$E = (C_{n} + C_{1} + C_{2} + \dots + C_{n}) + 2 (C_{1} + 2 C_{2} + \dots n C_{n}) = 2^{n} + 2 \cdot n 2^{n - 1} = (n + 1) 2^{n}$

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If (1+x)n=C0+C1x+C2x2+…+Cnxn, then C0+3C1+5C2+…+(2n+1)Cn=

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If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$ , then $C_{0} + 3 C_{1} + 5 C_{2} + \dots + (2 n + 1) C_{n} =$