If n is an odd natural number- then - r -0 n -1 r - r C r equals-A- 1 - nB- n - 2 nC- 0D- None

The correct option is C 0 Let n = 2k+1 then ∑^n_r=0( 1)^r/^nC_r = ∑^2k+1_r=0( 1)^2k+1/^2k+1C_r = ∑^n_r=0( 1)^r/^nC_r = ∑^2k+1_r=0( 1)^2k+1/^2k+1C_r = 1/^2k+1C ...

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Standard XII

Mathematics

Complex Numbers

If n is an od...

Question

If n is an odd natural number , then $n \sum r = 0 \frac{(- 1)^{r}}{^{r} C_{r}}$ equals :

1/n

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n/2ⁿ

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None

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Solution

The correct option is C
0

Let n = 2k+1 then $n \sum r = 0 \frac{(- 1)^{r}}{^{n} C_{r}} = 2 k + 1 \sum r = 0 \frac{(- 1)^{2 k + 1}}{^{2 k + 1} C_{r}}$

= $n \sum r = 0 \frac{(- 1)^{r}}{^{n} C_{r}} = 2 k + 1 \sum r = 0 \frac{(- 1)^{2 k + 1}}{^{2 k + 1} C_{r}} = \frac{1}{^{2 k + 1} C_{r}} - \frac{1}{^{2 k + 1} C_{1}} + . . . . + \frac{1}{^{2 k + 1} C_{2 k}} - \frac{1}{^{2 k + 1} C_{2 k + 1}} = 0$

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If n is an odd natural number , then n∑r=0(−1)rrCr equals :

The correct option is C 0 Let n = 2k+1 then n∑r=0(−1)rnCr=2k+1∑r=0(−1)2k+12k+1Cr = n∑r=0(−1)rnCr=2k+1∑r=0(−1)2k+12k+1Cr=12k+1Cr−12k+1C1+....+12k+1C2k−12k+1C2k+1=0

If n is an odd natural number , then $n \sum r = 0 \frac{(- 1)^{r}}{^{r} C_{r}}$ equals :