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Question: If (1 + x)<sup>n</sup> = C<sub>0</sub> + C<sub>1</sub>x + C<sub>2</sub>x<sup>2</sup> + ……+ C<sub>n</...

If (1 + x)n = C0 + C1x + C2x2 + ……+ Cnxn then

C01\frac{C_{0}}{1}+ C12\frac{C_{1}}{2}+ C23\frac{C_{2}}{3}+…… + Cnn+1\frac{C_{n}}{n + 1}=

A

2n+1

B

2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1}

C

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

D

None of these

Answer

2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1}

Explanation

Solution

Integrating :

C01\frac{C_{0}}{1}x + C12\frac{C_{1}}{2}x2 …..+Cnn+1\frac{C_{n}}{n + 1}xn+1 =(1+x)n+1−1n+1\frac{(1 + x)^{n + 1} - 1}{n + 1}

Put x = 1 ŽC01\frac{C_{0}}{1}+ C12\frac{C_{1}}{2}+……+Cnn+1\frac{C_{n}}{n + 1}= 2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1}