- Home
- Standard 11
- Mathematics
7.Binomial Theorem
medium
If ${x^4}$ occurs in the ${r^{th}}$ term in the expansion of ${\left( {{x^4} + \frac{1}{{{x^3}}}} \right)^{15}}$, then $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