JEE Main & Advanced Mathematics Binomial Theorem and Mathematical Induction Question Bank Expansion of binomial theorem

  • question_answer
    \[{{(\sqrt{2}+1)}^{6}}-{{(\sqrt{2}-1)}^{6}}=\] [MP PET 1984]

    A) 101

    B) \[70\sqrt{2}\]

    C) \[140\sqrt{2}\]

    D) \[120\sqrt{2}\]

    Correct Answer: C

    Solution :

    \[{{(x+a)}^{n}}-{{(x-a)}^{n}}\] \[=2\,{{[}^{n}}{{C}_{1}}{{x}^{n-1}}a+{{\,}^{n}}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+{{\,}^{n}}{{C}_{5}}{{x}^{n-5}}{{a}^{5}}+......]\] \ \[{{(\sqrt{2}+1)}^{6}}-{{(\sqrt{2}-1)}^{6}}\] \[=2\,{{[}^{6}}{{C}_{1}}{{(\sqrt{2})}^{5}}{{(1)}^{1}}+{{\,}^{6}}{{C}_{3}}{{(\sqrt{2})}^{3}}{{(1)}^{3}}+{{\,}^{6}}{{C}_{5}}{{(\sqrt{2})}^{1}}{{(1)}^{5}}]\] \[\because \,\,\,{{(\sqrt{2}+1)}^{6}}-{{(\sqrt{2}-1)}^{6}}=2[6\times 4\sqrt{2}+20\times 2\sqrt{2}+6\sqrt{2}]\] =\[2[24\sqrt{2}+40\sqrt{2}+6\sqrt{2}]=140\sqrt{2}\].


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