If n is an odd natural number- then -r-0n-1r- nCr equal to

The correct option is D 0 ∑_r=0 ^n( 1)^r/ ^nC_r =1 1/ ^nC_1+1/ ^nC_2 1/ ^nC_3+... 1/ ^nC_n ...(since n is odd, the ending term will have ...

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Sum of Product of Binomial Coefficients

If n is an ...

Question

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

0

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\frac{1}{n}

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

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

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If n is an even natural number , then $n \sum r = 0 \frac{(- 1)^{r}}{^{n} C_{r}} e q u a l s :$

n

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

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

f (x) = x^{n}

f (1) + \frac{f^{'} (1)}{1} + \frac{f^{''} (1)}{2!} + \frac{f^{'''} (1)}{3!} + . . . + \frac{^{n} f (1)}{n!} = 2^{n}

{(1 + x)}^{n} = n \sum r = 0^{n} C_{r} x^{r}

n \sum r = 0^{n} C_{r} = 2^{n}

n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{1}{2^{r}} + \frac{3^{r}}{2^{2 r}} + \frac{7^{r}}{2^{3 r}} + \dots \infty]

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