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If $n\in N$, then the value of $\sum_{r=0} ^{n} (-1)^r \large\frac{^nC_r}{^{r+2}C_r}$ is

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  • \[\sum_{r=0} ^{n+2} (-1)^ r. ^{n+2}C_{r}=^{n+2}C_0-^{n+2}C_1+^{n+2}C_2-......=(1-1)^{n+2}=0\]
  • $^{n+2}C_0=1\:\:and\:\:^{n+2}C_1=n+2$
$\large\frac{^nC_r}{{r+2}C_r}=$$\frac{n!}{(n-r)!.r!}.\frac{r!.2!}{(r+2)!}$
$=\large\frac{2}{(n+1).(n+2)}.\frac{(n+2)!}{(r+2)!.(n-r)!}$
$=\large\frac{2}{(n+1)(n+2)}$$.^{n+2}C_{r+2}$
\[\therefore\sum_{r=0} ^{n} (-1)^r.\frac{^nC_r}{{r+2}C_r}=\frac{2}{(n+1)(n+2)}.\sum_{r=0} ^{n} (-1)^ r. ^{n+2}C_{r+2}\]
\[=\frac{2}{(n+1)(n+2)}.\bigg[\sum_{r=0} ^{n+2} (-1)^ r. ^{n+2}C_{r}-^{n+2}C_0+^{n+2}C_1\bigg]\]
$=\large\frac{2}{(n+1)(n+2)}.$$[-1+(n+2)]$
$=\large\frac{2}{n+2}$
answered Oct 15, 2013 by rvidyagovindarajan_1
 

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