If coefficient of x in expansion of {left( {{x^2} + frac{k}{x}} right)^5} is 270 , then k =

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7.Binomial Theorem

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

A

$1$

B

$2$

C

$3$

D

$4$

Solution

(c) ${T_{r + 1}} = {}^5{C_r}{({x^2})^{5 – r}}{\left( {\frac{k}{x}} \right)^r}$

For coefficient of $x$, $10 – 2r – r = 1\,\,\,\, \Rightarrow r = 3$

Hence, ${T_{3 + 1}} = {}^5{C_3}{({x^2})^{5 – 3}}{\left( {\frac{k}{x}} \right)^3}$

According to question, $10\,{k^3} = 270\,\, \Rightarrow \,\,k = 3$.

Standard 11

Mathematics

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