B] \[{{\left( x+\FRAC{1}{x}+{{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{15}}={{\left( \frac{{{x}^{3}}+x+{{x}^{4}}+1}{{{x}^{2}}} \right)}^{15}}\] \[=\frac{{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{60}}{{x}^{60}}}{{{x}^{30}}}\] HENCE, the total number of terms is 61.

"> B] \[{{\left( x+\FRAC{1}{x}+{{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{15}}={{\left( \frac{{{x}^{3}}+x+{{x}^{4}}+1}{{{x}^{2}}} \right)}^{15}}\] \[=\frac{{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{60}}{{x}^{60}}}{{{x}^{30}}}\] HENCE, the total number of terms is 61.

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The number of distinct terms in the expansion of \[{{\left( x+\frac{1}{x}+{{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{15}}\]is/are (with respect to different power of x)

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 11 months ago

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[B] \[{{\left( x+\FRAC{1}{x}+{{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{15}}={{\left( \frac{{{x}^{3}}+x+{{x}^{4}}+1}{{{x}^{2}}} \right)}^{15}}\] \[=\frac{{{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+...+{{a}_{60}}{{x}^{60}}}{{{x}^{30}}}\] HENCE, the total number of terms is 61.

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