If the coefficient of x in the expansion of ( x 2 + k x ) 5 is 270, then k = Options: A . &nbsp 1 B . &nbsp 2 C . &nbsp 3 D . &nbsp 4

Answer: Option C : C T r + 1 = 5 C r ( x 2 ) 5 − r ( k x ) r For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3 Hence, T 3 + 1 = 5 C 3 ( x 2 ) 5 − 3 ( k x ) 3 According to question, 10 k 3 = 270 ⇒ k = 3.

If the coefficient of x in the expansion of (x2+kx)5 is 270, then k = &

Discussion Forum : Binomial Theorem

Question -

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

Options:

A . 1

B . 2

C . 3

D . 4

Answer: Option C
:
C

$T_{r + 1}$ = $^{5} C_{r}$ $(x^{2})^{5 - r}$ ( $\frac{k}{x})^{r}$

For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3

Hence, $T_{3 + 1}$ = $^{5} C_{3}$ $(x^{2})^{5 - 3}$ ( $\frac{k}{x})^{3}$

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

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