The term independent of x in the expansion of $\left(x+\frac{1}{x^2}\right)^6$ is

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Professor PC Thomas Classes · Answer

General term $T_{r+1} = { }^6C_r x^{6-r} (x^{-2})^r = { }^6C_r x^{6-3r}$. For independent term, $6-3r=0 \implies r=2$. Coefficient is ${ }^6C_2 = 15$.