Question

$\ \displaystyle \sf\frac{ \sum \limits _{r = 0}^{n} ( - 1) {}^{r} . {}^{n} C_r \bigg( \frac{1}{r + p + 1 } \bigg) }{\sum \limits _{r = 0}^{p} ( - 1) {}^{r} . {}^{p} C_r \bigg( \frac{1}{r + n + 1 } \bigg)} = \frac{1}{k}r=0∑p​(−1)r.pCr​(r+n+11​)\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater$
Find The Value Of 'k'

Don't Dare To Spam:))​

Answer 1

$\large\underline{\sf{Solution-}}$

Before we start the solution, Let's recall the propertie of Beta function to be used.

$\boxed{\tt{ B(m,n) = \displaystyle\int_{0}^{1}\rm {x}^{m} {(1 - x)}^{n} \: dx \: = B(n,m)}}$

Now,

We know, From Binomial Theorem, we have

$\rm :\longmapsto\: {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r}$

can be further rewritten as

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = {x}^{p} \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r}$

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r} \times {x}^{p}$

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r + p}$

On integrating both sides w. r. t. x, between x = 0 and x = 1, we get

$\rm :\longmapsto\:\displaystyle\int_{0}^{1}\rm {x}^{p} {(1 - x)}^{n}dx =\displaystyle\int_{0}^{1}\rm \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r + p}dx$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \displaystyle\int_{0}^{1}\rm {x}^{r + p}dx$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{ {x}^{r + p + 1} }{r + p + 1}\bigg]_{0}^{1}$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{1}{r + p + 1} - 0\bigg]$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{1}{r + p + 1} \bigg] - - (1)$

Similarly,

$\rm :\longmapsto\: B(n,p) = \displaystyle\sum_{r = 0}^{p}\rm {( - 1)}^{r} \: ^pC_r \: \bigg[\frac{1}{r + p + 1} \bigg] - - (2)$

Now, Consider

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

$\rm :\longmapsto\:\dfrac{B(n,p)}{B(p,n)} = \dfrac{1}{k}$

$\rm :\longmapsto\:\dfrac{1}{k} = 1$

$\bf\implies \:k \: = \: 1$

Answer 2

$\large\underline{\sf{Solution-}}$

Before we start the solution, Let's recall the propertie of Beta function to be used.

$\boxed{\tt{ B(m,n) = \displaystyle\int_{0}^{1}\rm {x}^{m} {(1 - x)}^{n} \: dx \: = B(n,m)}}$

Now,

We know, From Binomial Theorem, we have

$\rm :\longmapsto\: {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r}$

can be further rewritten as

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = {x}^{p} \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r}$

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r} \times {x}^{p}$

$\rm :\longmapsto\: {x}^{p} {(1 - x)}^{n} = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r + p}$

On integrating both sides w. r. t. x, between x = 0 and x = 1, we get

$\rm :\longmapsto\:\displaystyle\int_{0}^{1}\rm {x}^{p} {(1 - x)}^{n}dx =\displaystyle\int_{0}^{1}\rm \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: {x}^{r + p}dx$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \displaystyle\int_{0}^{1}\rm {x}^{r + p}dx$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{ {x}^{r + p + 1} }{r + p + 1}\bigg]_{0}^{1}$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{1}{r + p + 1} - 0\bigg]$

$\rm :\longmapsto\: B(p,n) = \displaystyle\sum_{r = 0}^{n}\rm {( - 1)}^{r} \: ^nC_r \: \bigg[\frac{1}{r + p + 1} \bigg] - - (1)$

Similarly,

$\rm :\longmapsto\: B(n,p) = \displaystyle\sum_{r = 0}^{p}\rm {( - 1)}^{r} \: ^pC_r \: \bigg[\frac{1}{r + p + 1} \bigg] - - (2)$

Now, Consider

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

$\rm :\longmapsto\:\dfrac{B(n,p)}{B(p,n)} = \dfrac{1}{k}$

$\rm :\longmapsto\:\dfrac{1}{k} = 1$

$\bf\implies \:k \: = \: 1$