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

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(C) The general term in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is given by:$T_{r+1} = { }^5 C_r (x^2)^{5-r} (\frac{k}{x})^r$$T_{r+1} = { }^5 C_r x^{10-2r} \cdot k^r \cdot x^{-r} = { }^5 C_r k^r x^{10-3r}$For the coefficient of $x$,we set the exponent of $x$ to $1$:$10 - 3r = 1$$3r = 9 \Rightarrow r = 3$Substituting $r = 3$ into the expression for the coefficient:Coefficient $= { }^5 C_3 k^3 = 10 k^3$Given that the coefficient is $270$:$10 k^3 = 270$$k^3 = 27$$k = 3$

Class 11 Mathematics Binomial Theorem General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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If the coefficient of $x$ in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is $270$,then $k$ is equal to

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$1$
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$2$
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$3$
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$4$

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Binomial Theorem

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General term, Coefficient of any power of x, Independent term, Middle term and Greatest term and Greatest coefficient

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If the coefficient of $x$ in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is $270$,then $k$ is equal to

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