The coefficient of {x^{32}} in the expansion | vedclass

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Question

(a) ${T_{r + 1}} = {}^{15}{C_r}{({x^4})^{15 - r}}{\left( {\frac{{ - 1}}{{{x^3}}}} ight)^r}$

$	herefore {T_{r + 1}} = {}^{15}{C_r}\frac{{{{(

Vedclass · Accepted Answer

$^{15}{C_4}$

Vedclass · Answer

(a) ${T_{r + 1}} = {}^{15}{C_r}{({x^4})^{15 - r}}{\left( {\frac{{ - 1}}{{{x^3}}}} ight)^r}$

$	herefore {T_{r + 1}} = {}^{15}{C_r}\frac{{{{(x)}^{60 - 4r}}{{( - 1)}^r}}}{{{{(x)}^{3r}}}}$

$ = {}^{15}{C_r}{( - 1)^r}{(x)^{60 - 7r}}$

Now putting $60 - 7r = 32$

==> $60 - 32 = 7r$

==> $r = \frac{{28}}{7} = 4$

$	herefore $ Coefficient of ${r^{32}} = {}^{15}{C_4}{( - 1)^4} = {}^{15}{C_4}$.

The coefficient of ${x^{32}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is

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