Question

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

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Answer 1

`\begin{gathered}(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 \ (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)\end{gathered}`

So required value

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