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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
$(25)^2$
$2^{25} -1$
$2^{24}$
$2^{25}$
Solution
$\sum\limits_{r = 1}^{25} {\frac{{\left| {50} \right.}}{{\left| r \right.\left| {50} \right. – r}} \times \frac{{\left| {50 – r} \right.}}{{\left| {25 – r\left| {25} \right.} \right.}}} $
$ = \sum\limits_{r = 1}^{25} {\frac{{\left| {50} \right.}}{{\left| {r\left| {25 – r\left| {25} \right.} \right.} \right.}}} $
$ = \frac{{\left| {50} \right.}}{{\left| {25} \right.}}\sum\limits_{r = 1}^{25} {\frac{1}{{\left| {r\left| {25 – r} \right.} \right.}}} $
$ = \frac{{\left| {50} \right.}}{{\left| {25\left| {25} \right.} \right.}}\sum\limits_{r = 1}^{25} {{\,^{25}}{C_r} = {\,^{50}}{C_{25}}} \left( {{2^{25}} – 1} \right)$
