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Question: \(\frac{C_{0}}{1} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + ..... + \frac{C_{n}}{n + 1} =\)...

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

A

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

B

2n−1n+1\frac{2^{n} - 1}{n + 1}

C

2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1}

D

None

Answer

2n+1−1n+1\frac{2^{n + 1} - 1}{n + 1}

Explanation

Solution

Consider the expansion

(1+x)n=C0+C1x+C2x2+....+Cnxn(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + .... + C_{n}x^{n}.....(i)

Integrating both sides of (i) within limits 0 to 1 we get, ∫01(1+x)ndx=∫01C0+∫01C1x+∫01C2x2+.......+∫01Cnxndx\int_{0}^{1}{(1 + x)^{n}}dx = \int_{0}^{1}{C_{0} + \int_{0}^{1}{C_{1}x} + \int_{0}^{1}{C_{2}x^{2} + ....... + \int_{0}^{1}{C_{n}x^{n}dx}}}

[(1+x)n+1n+1]01=C0[x]01+C1[x22]01+........+Cn[xn+1n+1]01\left\lbrack \frac{(1 + x)^{n + 1}}{n + 1} \right\rbrack_{0}^{1} = C_{0}\lbrack x\rbrack_{0}^{1} + C_{1}\left\lbrack \frac{x^{2}}{2} \right\rbrack_{0}^{1} + ........ + C_{n}\left\lbrack \frac{x^{n + 1}}{n + 1} \right\rbrack_{0}^{1}

2n+1n+1−1n+1=C0[1]+C112+C213+.......Cn.1n+1\frac{2^{n + 1}}{n + 1} - \frac{1}{n + 1} = C_{0}\lbrack 1\rbrack + C_{1}\frac{1}{2} + C_{2}\frac{1}{3} + .......C_{n}.\frac{1}{n + 1}; C0+C12+C23+....Cnn+1=2n+1−1n+1C_{0} + \frac{C_{1}}{2} + \frac{C_{2}}{3} + ....\frac{C_{n}}{n + 1} = \frac{2^{n + 1} - 1}{n + 1}.