First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

3
B

-3
C

9
D

-9

Solution

$T_{r + 1} =^{5} C_{r} {(x^{2})}^{5 - r} {(\frac{k}{x})}^{r} =^{5} C_{r} k^{r} x^{10 - 3 r}$
set $10 - 3 r = 1 o r r = 3 .$ Therefore,
$^{5} C_{3} k^{3} = 270 \Rightarrow k^{3} = 27 \Rightarrow k = 3$

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