A) \[{{(25)}^{2}}\]
B) \[{{2}^{25}}-1\]
C) \[{{2}^{24}}\]
D) \[{{2}^{25}}\]
Correct Answer: B
Solution :
| \[\sum\limits_{r=\,\,1}^{25}{\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 50-r \,}} \right. }}\times \frac{\left| \!{\underline {\, 50-r \,}} \right. }{\left| \!{\underline {\, 25-r \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }\] |
| \[=\sum\limits_{r=\,\,1}^{25}{\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 50-r \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }}=\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, 25 \,}} \right. }\sum\limits_{r=\,\,1}^{25}{\frac{1}{\left| \!{\underline {\, r \,}} \right. \,\left| \!{\underline {\, 25-r \,}} \right. }}\] |
| \[=\frac{\left| \!{\underline {\, 50 \,}} \right. }{\left| \!{\underline {\, 25 \,}} \right. \,\left| \!{\underline {\, 25 \,}} \right. }\sum\limits_{r=\,\,1}^{25}{{}^{25}{{C}_{r}}={}^{50}{{C}_{25}}}({{2}^{25}}-1).\] |
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