JCECE Engineering JCECE Engineering Solved Paper-2012

  • question_answer
    The coefficient of \[{{x}^{n}}\] in the expansion of \[{{(1-9x+20{{x}^{2}})}^{-1}}\]

    A) \[{{5}^{n}}-{{4}^{n}}\]                   

    B) \[{{5}^{n+1}}-{{4}^{n+1}}\]

    C) \[{{5}^{n-1}}-{{4}^{n-1}}\]                           

    D)  None of these

    Correct Answer: B

    Solution :

    We have, \[{{(1-9x+20{{x}^{2}})}^{-1}}={{[(1-5x)(1-4x)]}^{-1}}\]                 \[=\frac{1}{(1-5x)(1-4x)}=\frac{5}{1-5x}-\frac{4}{1-4x}\]                 \[=5{{(1-5x)}^{-1}}-4{{(1-4x)}^{-1}}\]                 \[=5[1+5x+{{(5x)}^{2}}+...+{{(5x)}^{n}}+...]\]                 \[-4[1+4x+{{(4x)}^{2}}+...+{{(4x)}^{n}}+...]\] \[\Rightarrow \]               The coefficient of\[{{x}^{n}}\] is\[{{5}^{n+1}}-{{4}^{n+1}}\]


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