7.Binomial Theorem
easy

The coefficient of ${x^{39}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is

A

$-455$

B

$-105$

C

$105$

D

$455$

Solution

(a) ${T_{r + 1}} = \,{}^{15}{C_r}{({x^4})^{15 – r}}{( – 1/{x^3})^r}$= ${( – 1)^r}\,\,{}^{15}{C_r}{(x)^{60 – 7r}}$

For coefficient of ${x^{39}}$, $60 – 7r = 39$

$\Rightarrow r = 3$

$\therefore {T_4} = {}^{15}{C_3}{({x^4})^{12}}{( – 1/{x^3})^3}$= $ – 455\,{x^{39}}$

Hence the required coefficient is $-455$.
Standard 11
Mathematics

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