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) \[a-{{\,}^{n}}{{C}_{1}}(a-1)+{{\,}^{n}}{{C}_{2}}(a-2)-...+{{(-1)}^{n}}(a-n)\] \[=a(0)+0=0\]
You need to login to perform this action.
You will be redirected in
3 sec
