C0−C12+C23−…+(−1)nCnn+1=

C0+C12x+C23x2+…+Cnn+1xn=(1+x)n+1−1(n+1)x⇒C0−C12+C23−C34+….+(−1)nCnn+1=−1(n+1)(−1)=1n+1.

C0−C12+C23−…+(−1)nCnn+1=

C0+C12x+C23x2+…+Cnn+1xn=(1+x)n+1−1(n+1)x⇒C0−C12+C23−C34+….+(−1)nCnn+1=−1(n+1)(−1)=1n+1.

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$C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \dots + (- 1)^{n} \frac{C_{n}}{n + 1} =$

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$\frac{2^{n + 1} - 1}{n + 1}$

$\frac{2^{n} - 1}{n + 1}$

$\frac{2^{n}}{n + 1}$

$\frac{1}{n + 1}$

answer is D.

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Detailed Solution

$\begin{array}{l} C_{0} + \frac{C_{1}}{2} x + \frac{C_{2}}{3} x^{2} + \dots + \frac{C_{n}}{n + 1} x^{n} = \frac{(1 + x)^{n + 1} - 1}{(n + 1) x} \\ \Rightarrow C_{0} - \frac{C_{1}}{2} + \frac{C_{2}}{3} - \frac{C_{3}}{4} + \dots . + (- 1)^{n} \frac{C_{n}}{n + 1} = \frac{- 1}{(n + 1) (- 1)} = \frac{1}{n + 1} . \end{array}$