C02+C13+C24+…+Cnn+2=

C0+C1x+C2x2+⋯+Cnxn=(1+x)n⇒C0x+C1x2+C2x3+⋯+Cnxn+1=x(1+x)n⇒∫01 C0x+C1x2+⋯+Cnxn+1dx=∫01 x(1+x)n⇒C0x22+C1x33+⋯+Cnxn+2n+201=x(1+x)n+1n+101−∫01 (1+x)n+1n+1dx⇒C02+C13+⋯+Cnn+2=2n+1n+1−(1+x)n+2(n+1)(n+2)01=2n+1n+1−2n+2−1(n+1)(n+2) =2n+1(n+2)−2n+2+1(n+1)(n+2)=n2n+1+1(n+1)(n+2).

C02+C13+C24+…+Cnn+2=

C0+C1x+C2x2+⋯+Cnxn=(1+x)n⇒C0x+C1x2+C2x3+⋯+Cnxn+1=x(1+x)n⇒∫01 C0x+C1x2+⋯+Cnxn+1dx=∫01 x(1+x)n⇒C0x22+C1x33+⋯+Cnxn+2n+201=x(1+x)n+1n+101−∫01 (1+x)n+1n+1dx⇒C02+C13+⋯+Cnn+2=2n+1n+1−(1+x)n+2(n+1)(n+2)01=2n+1n+1−2n+2−1(n+1)(n+2) =2n+1(n+2)−2n+2+1(n+1)(n+2)=n2n+1+1(n+1)(n+2).

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$\frac{C_{0}}{2} + \frac{C_{1}}{3} + \frac{C_{2}}{4} + \dots + \frac{C_{n}}{n + 2} =$

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$\frac{n \cdot 2^{n} + 5}{n}$

none

$\frac{n \cdot 2^{n} + 1}{(n + 1) (n + 2)}$

$\frac{n \cdot 2^{n + 1} + 1}{(n + 1) (n + 2)}$

answer is C.

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Detailed Solution

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