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JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Self Evaluation Test - Binomial Theorem

  • question_answer
    The sum of the series\[^{20}{{C}_{0}}-{{\,}^{20}}{{C}_{1}}+{{\,}^{20}}{{C}_{2}}-{{\,}^{20}}{{C}_{3}}+....\] \[-....+{{\,}^{20}}{{C}_{10}}\] is

    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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