यदि sum_{ r =0}^{25}left{{ }^{50} C _{ r } cd | vedclass

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Question

$\sum\limits_{r = 1}^{25} {\frac{{\left| {50} ight.}}{{\left| r ight.\left| {50} ight. - r}} 	imes \frac{{\left| {50 - r} ight.}}{{\left| {25

Vedclass · Accepted Answer

$2^{25} -1$

Vedclass · Answer

$\sum\limits_{r = 1}^{25} {\frac{{\left| {50} ight.}}{{\left| r ight.\left| {50} ight. - r}} 	imes \frac{{\left| {50 - r} ight.}}{{\left| {25 - r\left| {25} ight.} ight.}}} $

$ = \sum\limits_{r = 1}^{25} {\frac{{\left| {50} ight.}}{{\left| {r\left| {25 - r\left| {25} ight.} ight.} ight.}}} $

$ = \frac{{\left| {50} ight.}}{{\left| {25} ight.}}\sum\limits_{r = 1}^{25} {\frac{1}{{\left| {r\left| {25 - r} ight.} ight.}}} $

$ = \frac{{\left| {50} ight.}}{{\left| {25\left| {25} ight.} ight.}}\sum\limits_{r = 1}^{25} {{\,^{25}}{C_r} = {\,^{50}}{C_{25}}} \left( {{2^{25}} - 1} ight)$

यदि $\sum_{ r =0}^{25}\left\{{ }^{50} C _{ r } \cdot{ }^{50- r } C _{25- r }\right\}= K \left({ }^{50} C _{25}\right)$ हो, तो $K$ का मान होगा

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