The value of 2 nC1 +4 nC3 +6 nC5 +... is

The rth term in the given expression is Tr=2r nC2r−1 Since r+11. nCr=n+11. n+1Cr+1 ∴Tr=2r nC2r−1=n+11. n+1C2r ∴2 nC1+4 nC3+6 nC5+....=n+11[ n+1C2+ n+1C4+...] =n+11[2n+1−1− nC0]=n+12n−1

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The value of { { _{ }^{ n }{ C } }_{ 1 }^{ } }{ 2 } + { {

Question

The value of $\frac{_{n} C _{1}}{2} + \frac{_{n} C _{3}}{4} + \frac{_{n} C _{5}}{6} + ...$ is

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

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

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

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

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Text solutionVerified

The

r t h

term in the given expression is

T_{r} = \frac{_{n} C _{2 r - 1}}{2 r}

Since

\frac{1}{r + 1} ._{n} C_{r} = \frac{1}{n + 1} ._{n + 1} C_{r + 1}

∴ T_{r} = \frac{_{n} C _{2 r - 1}}{2 r} = \frac{1}{n + 1} ._{n + 1} C_{2 r}

∴ \frac{_{n} C _{1}}{2} + \frac{_{n} C _{3}}{4} + \frac{_{n} C _{5}}{6} + .... = \frac{1}{n + 1} [_{n + 1} C_{2} +_{n + 1} C_{4} + ...]

= \frac{1}{n + 1} [2^{n + 1 - 1} -_{n} C_{0}] = \frac{2 ^{n} - 1}{n + 1}

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Question Text	The value of $\frac{_{n} C _{1}}{2} + \frac{_{n} C _{3}}{4} + \frac{_{n} C _{5}}{6} + ...$ is
Answer Type	Text solution:1
Upvotes	150

The value of 2 n​C1 ​​+4 n​C3 ​​+6 n​C5 ​​+... is

Text solutionVerified

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The value of $\frac{_{n} C _{1}}{2} + \frac{_{n} C _{3}}{4} + \frac{_{n} C _{5}}{6} + ...$ is