Coefficient of x in the expansion of {left( { | vedclass

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(B) The general term in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is given by ${T_{r + 1}} = {\,^5}{C_r}{({x^2})^{5 - r}}{\left( {\f

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$10{a^3}$

Vedclass · Answer

(B) The general term in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is given by ${T_{r + 1}} = {\,^5}{C_r}{({x^2})^{5 - r}}{\left( {\frac{a}{x}} \right)^r}$.Simplifying this,we get ${T_{r + 1}} = {\,^5}{C_r}{a^r}{x^{10 - 2r}}{x^{-r}} = {\,^5}{C_r}{a^r}{x^{10 - 3r}}$.To find the coefficient of $x$,we set the exponent of $x$ equal to $1$:$10 - 3r = 1$$3r = 9$$r = 3$.Substituting $r = 3$ into the general term:${T_{3 + 1}} = {\,^5}{C_3}{a^3}{x^{10 - 3(3)}}$${T_4} = 10 \cdot {a^3} \cdot x$.Thus,the coefficient of $x$ is $10{a^3}$.

Coefficient of $x$ in the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ is

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