A) 0
B) \[^{20}{{C}_{10}}\]
C) \[{{-}^{20}}{{C}_{10}}\]
D) \[\frac{1}{2}{{\,}^{20}}{{C}_{10}}\]
Correct Answer: D
Solution :
[d] We know that, \[{{(1+x)}^{20}}={{\,}^{20}}{{C}_{0}}+{{\,}^{20}}{{C}_{1}}x+\] |
\[^{20}{{C}_{2}}{{x}^{2}}+....{{\,}^{20}}{{C}_{10}}{{x}^{10}}+....{{\,}^{20}}{{C}_{20}}{{x}^{20}}\] |
Put \[x=-1,\,(0)={{\,}^{20}}{{C}_{0}}-{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}-{{\,}^{20}}{{C}_{3}}+....\] |
\[{{+}^{20}}{{C}_{10}}-{{\,}^{20}}{{C}_{11}}...+{{\,}^{20}}{{C}_{20}}\] |
\[\Rightarrow 0=2{{[}^{20}}{{C}_{0}}-{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}-{{\,}^{20}}{{C}_{3}}+....-{{\,}^{20}}{{C}_{9}}]\] |
\[+{{\,}^{20}}{{C}_{10}}\] |
\[\Rightarrow {{\,}^{20}}{{C}_{10}}=2{{[}^{20}}{{C}_{0}}-{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}-{{\,}^{20}}{{C}_{3}}\] |
\[+....-{{\,}^{20}}{{C}_{9}}+{{\,}^{20}}{{C}_{10}}]\] |
\[\Rightarrow {{\,}^{20}}{{C}_{0}}-{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}-{{\,}^{20}}{{C}_{3}}+...+{{\,}^{20}}{{C}_{10}}\] |
\[=\frac{1}{2}{{\,}^{20}}{{C}_{10}}\] |
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