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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