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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 28

  • question_answer
    The value of  \[^{n}{{C}_{1}}\left( \sum\limits_{r=0}^{1}{^{1}{{C}_{r}}} \right){{+}^{n}}{{C}_{2}}\left( \sum\limits_{r=0}^{2}{^{2}{{C}_{r}}} \right){{+}^{n}}{{C}_{3}}\left( \sum\limits_{r=0}^{3}{^{3}{{C}_{r}}} \right)+...{{+}^{n}}{{C}_{n}}\left( \sum\limits_{r=0}^{n}{^{n}{{C}_{r}}} \right)\]is equal to

    A) \[{{2}^{n}}\]                  

    B)        \[{{3}^{n}}\]                  

    C) \[{{3}^{n}}-1\]    

    D)        \[{{3}^{n}}+1\]

    Correct Answer: C

    Solution :

    [c] Given series is \[^{n}{{C}_{1}}\times {{2}^{1}}{{+}^{n}}{{C}_{2}}\times {{2}^{2}}{{+}^{n}}{{C}_{3}}\times {{2}^{3}}+....{{+}^{n}}{{C}_{n}}\times {{2}^{n}}\] \[=[1{{+}^{n}}{{C}_{1}}\times {{2}^{1}}{{+}^{n}}{{C}_{2}}\times {{2}^{2}}{{+}^{n}}{{C}_{3}}\times {{2}^{3}}+....{{+}^{n}}{{C}_{n}}\times {{2}^{n}})-1\]\[={{(1+2)}^{n}}-1={{3}^{n}}-1\]          


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