The coefficient of {x^{39}} in the expansion | vedclass

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Question

(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 ${x^{39}}$, $6

Vedclass · Accepted Answer

$-455$

Vedclass · Answer

(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 ${x^{39}}$, $60 - 7r = 39$

$\Rightarrow r = 3$

$	herefore {T_4} = {}^{15}{C_3}{({x^4})^{12}}{( - 1/{x^3})^3}$= $ - 455\,{x^{39}}$

Hence the required coefficient is $-455$.

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

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