2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    if\[|x|<1,\]then the coefficient of\[{{x}^{n}}\]in \[{{(1+2x+3{{x}^{2}}+4{{x}^{3}}+....)}^{1/2}},\]is:

    A)  n                                           

    B) \[n+1\]

    C)  \[-n\]                  

    D)         \[-1\]

    E)  1

    Correct Answer: E

    Solution :

    \[{{(1+2x+3{{x}^{2}}+.....)}^{1/2}}\] \[={{[{{(1-x)}^{2}}]}^{1/2}}={{(1-x)}^{-1}}\] \[=1+x+{{x}^{2}}+....+{{x}^{n}}+...\]        \[\therefore \]Coefficient of\[{{x}^{n}}=1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner