If n is odd natural number, then ∑r=0n nCr(−1) r equals................

r=0∑nnCr(−1) r =nC01−nC11+nC21+.......+nCn−11−nCn1 (Since n is odd natural number)r=0∑nnCr(−1) r =nCn1−nCn−11+nCn−21+.......+nCn−11−nCn1 (∵nCr=nCn−r )=0

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If n is odd natural number, then _{ r=0 }^{ n }{ { { ( -1 )

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If $n$ is odd natural number, then $\sum_{r = 0}^{n} \frac{( - 1 ) ^{r}}{_{n} C _{r}}$ equals................

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Michael Discussed

n

is odd natural number, then

\sum_{r = 0}^{n} \frac{( - 1 ) ^{r}}{_{n} C _{r}}

equals................

14 mins ago

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14 mins ago

Text solutionVerified

r = 0 \sum n \frac{( - 1 ) ^{r}}{^{n} C _{r}} = \frac{1}{^{n} C _{0}} - \frac{1}{^{n} C _{1}} + \frac{1}{^{n} C _{2}} + ....... + \frac{1}{^{n} C _{n - 1}} - \frac{1}{^{n} C _{n}}

(Since

n

is odd natural number)

r = 0 \sum n \frac{( - 1 ) ^{r}}{^{n} C _{r}} = \frac{1}{^{n} C _{n}} - \frac{1}{^{n} C _{n - 1}} + \frac{1}{^{n} C _{n - 2}} + ....... + \frac{1}{^{n} C _{n - 1}} - \frac{1}{^{n} C _{n}}

(

∵^{n} C_{r} =^{n} C_{n - r}

)

= 0

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Question Text	If $n$ is odd natural number, then $\sum_{r = 0}^{n} \frac{( - 1 ) ^{r}}{_{n} C _{r}}$ equals................
Answer Type	Text solution:1
Upvotes	150

If n is odd natural number, then ∑r=0n​ n​Cr​(−1) r​ equals................

Text solutionVerified

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