If sumlimits_{r = 0}^{25} {left{ {^{50}{ | 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)$

If $\sum\limits_{r = 0}^{25} {\left\{ {^{50}{C_r}.{\,^{50 - r}}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} $, then $K$ is equal to

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