A) \[{{2}^{19}}+\frac{1}{2}{{\,}^{20}}{{C}_{10}}\]
B) \[{{2}^{19}}\]
C) \[^{20}{{C}_{10}}\]
D) None of these
Correct Answer: A
Solution :
\[\sum\limits_{K=0}^{10}{^{20}{{C}_{k}}}\] i.e., \[^{20}{{C}_{0}}+{{\,}^{20}}{{C}_{1}}+......+{{\,}^{20}}{{C}_{10}}\] We know that, \[{{(1+x)}^{n}}={{\,}^{n}}{{C}_{0}}+{{\,}^{n}}{{C}_{1}}{{x}^{1}}+{{\,}^{n}}{{C}_{2}}{{x}^{2}}+....+{{\,}^{n}}{{C}_{n}}.{{x}^{n}}\] Put\[x=1\]; \[{{2}^{n}}={{\,}^{n}}{{C}_{0}}+{{\,}^{n}}{{C}_{1}}+{{\,}^{n}}{{C}_{2}}+.....+{{\,}^{n}}{{C}_{n}}\] Put\[n=20\]; \[{{2}^{20}}={{\,}^{20}}{{C}_{0}}+{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}+......+{{\,}^{20}}{{C}_{20}}\] \[{{2}^{20}}+\,{{\,}^{20}}{{C}_{10}}=2\,[{{\,}^{20}}{{C}_{0}}+{{\,}^{20}}{{C}_{1}}+......+{{\,}^{20}}{{C}_{10}}]\] \[{{[}^{20}}{{C}_{0}}+{{\,}^{20}}{{C}_{1}}+......+{{\,}^{20}}{{C}_{10}}]={{2}^{19}}+\frac{1}{2}{{\,}^{20}}{{C}_{10}}\] \[\sum\limits_{k=0}^{10}{^{20}{{C}_{k}}}={{2}^{19}}+\frac{1}{2}{{\,}^{20}}{{C}_{10}}\].
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