A) \[{{b}_{n}}={{b}_{n-1}}={{a}_{n}}\]
B) \[{{b}_{n}}-{{b}_{n-1}}={{a}_{n}}\]
C) \[{{b}_{n}}/{{b}_{n-1}}={{a}_{n}}\]
D) none of these
Correct Answer: B
Solution :
[b] \[\frac{f(x)}{1-x}={{b}_{0}}+{{b}_{1}}x+{{b}_{2}}{{x}^{2}}+...+{{b}_{n}}{{x}^{n}}+...\] \[\Rightarrow {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{n}}{{x}^{n}}+...\] \[=(1-x)({{b}_{0}}+{{b}_{1}}x+{{b}_{2}}{{x}^{2}}+...+{{b}_{n}}{{x}^{n}}+...)\] Comparing the coefficient of \[{{x}^{n}}\]on both the sides, \[{{a}_{n}}={{b}_{n}}-{{b}_{n-1.}}\]
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