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

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Question

(b) In the expansion of ${\left( {{x^2} + \frac{a}{x}} \right)^5}$ the general term is ${T_{r + 1}} = {\,^5}{C_r}{({x^2})^{5 - r}}{\left( {\frac{a}{x}

Vedclass · Accepted Answer

$10{a^3}$

Vedclass · Answer

(b) In the expansion of ${\left( {{x^2} + \frac{a}{x}} ight)^5}$ the general term is ${T_{r + 1}} = {\,^5}{C_r}{({x^2})^{5 - r}}{\left( {\frac{a}{x}} ight)^r}$

$= {\,^5}{C_r}{a^r}{x^{10 - 3r}}$

Here, exponent of $x$ is $10 - 3r = 1$

$\Rightarrow r = 3$$	herefore $ ${T_{2 + 1}} = {\,^5}{C_3}{a^3}x = 10{a^3}.x$

Hence 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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