A) \[{{2}^{19}}+\frac{1}{2}\]\[^{20}{{C}_{10}}\]
B) \[{{2}^{19}}\]
C) \[^{20}{{C}_{10}}\]
D) None of these
Correct Answer: A
Solution :
\[\sum\limits_{k=0}^{10}{^{20}{{C}_{k}}{{=}^{20}}{{C}_{0}}{{+}^{20}}{{C}_{1}}+...{{+}^{20}}{{C}_{10}}}\] \[=\frac{1}{2}[2{{\cdot }^{20}}{{C}_{0}}+2{{\cdot }^{20}}{{C}_{1}}+...+2{{\cdot }^{20}}{{C}_{10}}]\] \[=\frac{1}{2}[{{(}^{20}}{{C}_{0}}{{+}^{20}}{{C}_{20}})]+{{(}^{20}}{{C}_{1}}{{+}^{20}}{{C}_{19}})+\] \[...+{{2}^{20}}{{C}_{10}}]\] \[=\frac{10}{2}[{{(}^{20}}{{C}_{0}}{{+}^{20}}{{C}_{1}}{{+}^{20}}{{C}_{2}}+...{{+}^{20}}{{C}_{19}}\] \[{{+}^{20}}{{C}_{20}}){{+}^{20}}{{C}_{10}}]\] \[=\frac{1}{2}[{{2}^{29}}{{+}^{20}}{{C}_{10}}]\] \[={{2}^{19}}+\frac{1}{2}\]\[^{20}{{C}_{10}}\]
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