If n is an even natural number - then -r-0n-1r-n Cr equals -

The correct option is D n is even . Let n = 2k, then nbsp; ∑_r=0^n( 1)^r/^nC_r = ∑_r=0^k 1( ( 1)^r/C_r + ( 1)^n r/C_n r) + ( 1)^k/C_k nbsp; = 2 ∑_r=0^k 1 ...

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Standard XI

Mathematics

Standard Limits to Remove Indeterminate Form

If n is an ev...

Question

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

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$2 \sum\limits_{r=0}^{\frac n2 - 1} \frac{(-1)^r}{{}^nC_r} + \frac{(-1)^{\frac n2}}{C_{\frac n2}}$

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Solution

The correct option is D
$2 \sum\limits_{r=0}^{\frac n2 - 1} \frac{(-1)^r}{{}^nC_r} + \frac{(-1)^{\frac n2}}{C_{\frac n2}}$

n is even . Let n = 2k, then

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

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

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

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Similar questions

Q. If

n

is an odd natural number, then

n \sum r = 0 \frac{(- 1)^{r}}{^{n} C_{r}}

equal to

Q. If

n

is odd natural number, then

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

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

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

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]

is equal to

S = n \sum r = 0 (- 1)^{r}^{n} C_{r} [\frac{2^{r}}{3^{r}} + \frac{8^{r}}{3^{2 r}} + \frac{26^{r}}{3^{3 r}} + . . . \infty]

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If n is an even natural number , then n∑r=0(−1)rnCrequals:

The correct option is D n is even . Let n = 2k, then n∑r=0(−1)rnCr=k−1∑r=0((−1)rCr+(−1)n−rCn−r)+(−1)kCk = 2k−1∑r=0(−1)rCr+(−1)kCk = 2n2−1∑r=0(−1)rnCr+(−1)n2Cn2

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