JEE Main & Advanced Sample Paper JEE Main Sample Paper-21

  • question_answer
    If the coefficient of \[{{x}^{100}}\]in \[1+(x+1)+{{(1+x)}^{2}}+.......+{{(1+x)}^{n}};(n\ge :100)\] is \[^{201}{{C}_{101}},\] then n is equal to

    A)  100                             

    B)  101

    C)  200                             

    D)  201

    Correct Answer: C

    Solution :

    \[^{100}{{C}_{100}}{{+}^{101}}{{C}_{100}}{{+}^{102}}{{C}_{100}}+{{..........}^{n}}{{C}_{100}}+\] \[\underbrace{\underbrace{^{101}{{C}_{101}}{{+}^{101}}{{C}_{100}}}_{^{102}{{C}_{101}}}\,\,{{\,}^{102}}{{C}_{100}}+}_{^{103}{{C}_{101}}}...........{{+}^{n}}{{C}_{100}}\] \[\therefore \,\,S{{=}^{n+1}}{{C}_{101}}\,{{=}^{201}}{{C}_{101}}\] \[\therefore \,n+1=201\,\Rightarrow \,n=200\]


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