Question

$\displaystyle \sf\frac{ \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \bigg( \frac{1}{r + p + 1 } \bigg) }{\sum \limits _{r = 0}^{p} ( - 1) {}^{r} . {}^{p} C_r \bigg( \frac{1}{r + n + 1 } \bigg)} = \frac{1}{k}$

Find the value of k​

Answer 1

Answer:

1

Step-by-step explanation:

$(1 - x) {}^{n} = { \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \: {x}^{r} } \\ \\ multiply \: both \: sides \: by \: {x}^{p} \\ \\ {x}^{p} (1 - x) {}^{n} = { \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \: {x}^{r + p} } \\ \\ integrate \: both \: sides \: \: limit \: 0 \: to \: 1\\ \\ \int _{0} ^{1} {x}^{p} (1 - x) {}^{n} \: dx = { \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \int_{0} ^{1} \: {x}^{r + p} } \: dx \\ \\ \int _{0} ^{1} {x}^{p} (1 - x) {}^{n} \: dx = { \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \ \frac{1}{r + p + 1} } \\ \\ \implies \sum \limits _{r = 0}^{n} \frac{( - 1) {}^{r}}{n + p+ 1} . {}^{n} C_r = \beta(n,p) \\ \\ simillarly \\ \\ \sum \limits _{r = 0}^{p} \frac{( - 1) {}^{r}}{r + n+ 1} . {}^{p} C_r = \beta(p,n)$

So required value

$\displaystyle \sf\frac{ \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \bigg( \frac{1}{r + p + 1 } \bigg) }{\sum \limits _{r = 0}^{p} ( - 1) {}^{r} . {}^{p} C_r \bigg( \frac{1}{r + n + 1 } \bigg)} = \frac{\beta(p,n)}{\beta(n,p)} \\ \\ since \: \beta(n,p) = \beta(p,n) \\ \\ \implies \frac{1}{k} = 1 \implies \: k = 1$