7.Binomial Theorem
medium

यदि ${\left( {{x^4} + \frac{1}{{{x^3}}}} \right)^{15}}$ के विस्तार में $r$ वें पद में ${x^4}$ आता है, तो $r = $ 

A

$7$

B

$8$

C

$9$

D

$10$

Solution

${T_r} = {\,^{15}}{C_{r – 1}}{({x^4})^{16 – r}}{\left( {\frac{1}{{{x^3}}}} \right)^{r – 1}} = {\,^{15}}{C_{r – 1}}{x^{67 – 7r}}$

 $67 – 7r = 4 $

$\Rightarrow r = 9$.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.