2. Solved Papers
  3. BCECE Engineering

BCECE Engineering BCECE Engineering Solved Paper-2004

  • question_answer
    If \[{{a}_{r}}\]is the coefficient of \[{{x}^{r}}\] in the expansion of \[{{(1+x+{{x}^{2}})}^{n}},\]then \[{{a}_{1}}-2{{a}_{2}}+3{{a}_{2}}-...-2n{{a}_{2n}}\]is equal to:    

    A)  0                            

    B)         \[n\]                    

    C) \[-n\]                   

    D)         \[2n\]

    Correct Answer: C

    Solution :

    Let us take \[{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{2n}}{{x}^{2n}}={{(1+x+{{x}^{2}})}^{n}}\] On differentiating w.r.t. \[x\] on both sides, we get \[{{a}_{1}}+2{{a}_{2}}x+...+2n{{a}_{2n}}{{x}^{2n-1}}\] \[=n{{(1+x+{{x}^{2}})}^{n-1}}(2x+1)\] Put \[x=-1\] \[\Rightarrow \]               \[{{a}_{1}}-2{{a}_{2}}+3{{a}_{3}}-...+2n{{a}_{2n}}=-n\]


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