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.
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