7.Binomial Theorem
medium

જો ${\left( {{x^4} + \frac{1}{{{x^3}}}} \right)^{15}}$ ના વિસ્તરણમાં ${x^4}$ એ ${r^{th}}$ પદમાં બને છે તો $r = $

A

$7$

B

$8$

C

$9$

D

$10$

Solution

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

==> $67 – 7r = 4 $

$\Rightarrow r = 9$.

Standard 11
Mathematics

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