A) -4
B) 6
C) -8
D) 10
Correct Answer: A
Solution :
\[1+{{x}^{4}}+{{x}^{5}}=\sum\limits_{i=0}^{5}{{{a}_{i}}{{(1+x)}^{i}}}\] \[={{a}_{0}}+{{a}_{1}}{{(1+x)}^{1}}+{{a}_{2}}{{(1+x)}^{2}}+{{a}_{3}}{{(1+x)}^{3}}\] \[+{{a}_{4}}{{(1+x)}^{4}}+{{a}_{5}}{{(1+x)}^{5}}\] \[\Rightarrow \]\[1+{{x}^{4}}+{{x}^{5}}\] \[={{a}_{0}}+{{a}_{1}}(1+x)+{{a}_{2}}(1+2x+{{x}^{2}})+{{a}_{3}}(1+3x+3{{x}^{2}}+{{x}^{3}})\]\[={{a}_{4}}(1+4x+6{{x}^{2}}+4{{x}^{3}}+{{x}^{4}})+{{a}_{5}}(1+5x+10{{x}^{2}}+10{{x}^{3}}+5{{x}^{4}}+{{x}^{5}})\]\[\Rightarrow \]\[1+{{x}^{4}}+{{x}^{5}}\] \[={{a}_{0}}+{{a}_{1}}+{{a}_{1}}x+{{a}_{2}}+2{{a}_{2}}x+{{a}_{2}}{{x}^{2}}+{{a}_{3}}+3{{a}_{3}}x\] \[+3{{a}_{3}}{{x}^{2}}+{{a}_{3}}{{x}^{3}}+{{a}_{4}}+4{{a}_{4}}x+6{{a}_{4}}{{x}^{2}}+4{{a}_{4}}{{x}^{3}}+{{a}_{4}}{{x}^{4}}+{{a}_{5}}\]\[+5{{a}_{5}}x+10{{a}_{5}}{{x}^{2}}+10{{a}_{5}}{{x}^{3}}+5{{a}_{5}}{{x}^{4}}+{{a}_{5}}{{x}^{5}}\] \[\Rightarrow \]\[1+{{x}^{4}}+{{x}^{5}}\] \[=({{a}_{0}}={{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+{{a}_{5}})+({{a}_{1}}+2{{a}_{2}}+3{{a}_{3}}+4{{a}_{4}}+5{{a}_{5}})\]\[+{{x}^{2}}({{a}_{2}}+3{{a}_{3}}+3{{a}_{3}}+6{{a}_{4}}+10{{a}_{5}})+{{x}^{3}}({{a}_{3}}+4{{a}_{4}}+10{{a}_{5}})\]\[={{x}^{4}}({{a}_{4}}+5{{a}_{5}})+{{x}^{5}}({{a}_{5}})\] On comparing the like coefficients, we get \[\]?(1); \[\]?.(2); \[\]?(3)and\[\]?.(4) from (1) & (2), we get \[\] ...(5) from (1), (3) & (5), we get \[\]...(6) Now, from (1), (5) and (6), we get \[{{a}_{2}}=-4\]
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