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If C_{0}, C_{1}, C_{2}, ...., C_{n} are binomial coefficients

Question

If $C_{0}, C_{1}, C_{2}, ...., C_{n}$ are binomial coefficients of order $n$ , then the value of $\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + .... =$

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

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

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

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

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Sophia Discussed

C_{0}, C_{1}, C_{2}, ...., C_{n}

are binomial coefficients of order

n

, then the value of

\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + .... =

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Text solutionVerified

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + ...... C_{n} x^{n}

Integrating, We have

\frac{( 1 + x ) ^{n + 1}}{n + 1} = C_{0} x + C_{1} \frac{x ^{2}}{2} + C_{2} \frac{x ^{3}}{3} + ..... C_{n} \frac{x ^{n + 1}}{n + 1}

Putting,

x = 1

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

Putting,

x = - 1

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

Adding above two equations we have,

\frac{( 2 ) ^{n + 1}}{n + 1} = 2 (\frac{C _{1}}{2} + \frac{C _{3}}{4} .....)

⟹ \frac{( 2 ) ^{n}}{n + 1} = \frac{C _{1}}{2} + \frac{C _{3}}{4} .....

Hence, option D is correct.

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Question Text	If $C_{0}, C_{1}, C_{2}, ...., C_{n}$ are binomial coefficients of order $n$ , then the value of $\frac{C _{1}}{2} + \frac{C _{3}}{4} + \frac{C _{5}}{6} + .... =$
Answer Type	Text solution:1
Upvotes	150