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Question: The value of \(\frac{C_{1}}{2} + \frac{C_{3}}{4} + \frac{C_{5}}{6} + .....\) is equal to...

The value of C12+C34+C56+.....\frac{C_{1}}{2} + \frac{C_{3}}{4} + \frac{C_{5}}{6} + ..... is equal to

A

2n−1n+1\frac{2^{n} - 1}{n + 1}

B

n.2nn.2^{n}

C

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

D

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

Answer

2n−1n+1\frac{2^{n} - 1}{n + 1}

Explanation

Solution

We have C12+C34+C56+......\frac{C_{1}}{2} + \frac{C_{3}}{4} + \frac{C_{5}}{6} + ......

= n2.1+n(n−1)(n−2)4.3.2.1+n(n−1)(n−2)(n−3)(n−4)6.5.4.3.2.1+.....\frac{n}{2.1} + \frac{n(n - 1)(n - 2)}{4.3.2.1} + \frac{n(n - 1)(n - 2)(n - 3)(n - 4)}{6.5.4.3.2.1} + .....

= 1n+1[(n+1)n2!+(n+1)(n)(n−1)(n−2)4!+.....]\frac{1}{n + 1}\left\lbrack \frac{(n + 1)n}{2!} + \frac{(n + 1)(n)(n - 1)(n - 2)}{4!} + ..... \right\rbrack

= 1n+1[2(n+1)−1−1]=2n−1n+1\frac{1}{n + 1}\lbrack 2^{(n + 1) - 1} - 1\rbrack = \frac{2^{n} - 1}{n + 1}

Trick: For n=1n = 1, = C12=1C12=12\frac{C_{1}}{2} = \frac{1C_{1}}{2} = \frac{1}{2}

Which is given by option (1) 2n−1n+1=21−11+1=12\frac{2^{n} - 1}{n + 1} = \frac{2^{1} - 1}{1 + 1} = \frac{1}{2}.