BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer If n is an integer greater than 1, then \[a-{{\,}^{n}}{{C}_{1}}(a-1)+{{\,}^{n}}{{C}_{2}}(a-2)-....+{{(-1)}^{n}}(a-n)\] is equal to:

    A) \[a\]                                     

    B)                  \[0\]                                     

    C)                  \[{{a}^{2}}\]                                      

    D)                  \[{{2}^{n}}\]

    Correct Answer: B

    Solution :

    \[a-{{\,}^{n}}{{C}_{1}}(a-1)+{{\,}^{n}}{{C}_{2}}(a-2)-\] \[....+{{(-1)}^{n}}(a-n)\]                 \[=a({{\,}^{n}}{{C}_{0}}-{{\,}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{-}^{n}}{{C}_{3}}+....+{{(-1)}^{n}}\,{{\,}^{n}}{{C}_{n}})\]                 \[+\,({{\,}^{n}}{{C}_{1}}-2{{\,}^{n}}{{C}_{2}}+3{{\,}^{n}}{{C}_{3}}-....+{{(-1)}^{n+1}}n{{\,}^{n}}{{C}_{n}})\]                                                                                 ?(i) As we know \[{{(1-x)}^{n}}={{\,}^{n}}{{C}_{0}}-{{x}^{n}}{{C}_{1}}+{{x}^{2n}}{{C}_{2}}-{{x}^{3}}{{\,}^{n}}C{{\,}_{3}}+\] \[....+{{(-1)}^{n}}{{x}^{n}}{{\,}^{n}}{{C}_{n}}\]                    ?(ii) Put \[x=1,\]we get \[0={{\,}^{n}}{{C}_{0}}={{\,}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}{{-}^{n}}{{C}_{3}}+...+{{(-1)}^{n}}{{C}_{n}}\] On differentiating Eq. (ii) w.r.t.\[x,\] we get \[n{{(1-x)}^{n-1}}=-{{\,}^{n}}{{C}_{1}}+2x{{\,}^{n}}{{C}_{2}}-3{{x}^{2}}{{\,}^{n}}{{C}_{3}}+...\] \[+{{(-1)}^{n}}n\,{{x}^{n-1}}{{\,}^{n}}{{C}_{n}}\]                 Put \[x=1\]                 \[0=-{{\,}^{n}}{{C}_{1}}+2{{\,}^{n}}{{C}_{2}}-3\,{{\,}^{n}}{{C}_{3}}+....+{{(-1)}^{n-1}}n{{\,}^{n}}{{C}_{n}}\]                 From Eq. (i)                 \[a-{{\,}^{n}}{{C}_{1}}(a-1)+{{\,}^{n}}{{C}_{2}}(a-2)-...+{{(-1)}^{n}}(a-n)\] \[=a(0)+0=0\]

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