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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    If \[^{20}{{C}_{1}}+{{\left( {{2}^{2}} \right)}^{20}}{{C}_{2}}+{{\left( {{3}^{2}} \right)}^{20}}{{C}_{3}}+.......+{{({{20}^{2}})}^{20}}{{C}_{20}}\]\[=A({{2}^{\beta }}),\] then the ordered pair \[(A,\beta )\]is equal to: [JEE Main 12-4-2019 Afternoon]

    A) (420, 18)

    B) (380, 19)

    C) (380, 18)                     

    D) (420, 19)

    Correct Answer: A

    Solution :

    \[{{(1+x)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+....+{}^{n}{{C}_{n}}{{x}^{n}}\]Diff. w.r.t. x           \[\Rightarrow \]\[n{{(1+x)}^{n-1}}={}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}(2x)+....+{}^{n}{{C}_{n}}n{{(x)}^{n-1}}\] Multiply by x both side \[\Rightarrow nx{{(1+x)}^{n-1}}={}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}(2{{x}^{2}})+....+{}^{n}{{C}_{n}}(n{{x}^{n}})\] Diff w.r.t. x \[\Rightarrow n[{{(1+x)}^{n-1}}\,+(n-1)x{{(1+x)}^{n-2}}]\] \[={}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}{{2}^{2}}x+{{....}^{n}}{{C}_{n}}({{n}^{2}}){{x}^{n-1}}\] Put x = 1 and n = 20 \[\Rightarrow {}^{20}{{C}_{1}}+{{2}^{2}}{{\,}^{20}}{{C}_{2}}+{{3}^{2}}{{\,}^{20}}{{C}_{3}}+....+{{(20)}^{2}}{{\,}^{20}}{{C}_{20}}\] \[=20\times {{2}^{18}}[2+19]=420({{2}^{18}})=A({{2}^{\beta }})\]                   


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