If the coefficient of x in the expansion of x2-k-x5 is 270- then k-A- 1B- 2C- 3D- 4

The correct option is C 3 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 ...

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Standard XI

Mathematics

Binomial Coefficients

If the coeffi...

Question

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

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Solution

The correct option is C
3

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

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

The correct option is C 3 Tr+1 = 5Cr(x2)5−r(kx)r For coefficient of x, 10 - 2r - r = 1 ⇒ r = 3 Hence, T3+1 = 5C3(x2)5−3(kx)3 According to question, 10k3 = 270 ⇒ k = 3.

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

The correct option is C
3

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

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.