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$C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + .. + \frac{C_{n}}{n+1} =$

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2n+1-1n+1

2n-1n+1

2nn+1

1n+1

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detailed solution

Correct option is A

C0+C12x+C23x2+…+Cnn+1xn=(1+x)n+1−1(n+1)x⇒C0+C12+C23+…+Cnn+1=2n+1−1n+1.

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C0+C12+C23+..+Cnn+1=

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$C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + .. + \frac{C_{n}}{n+1} =$