The sum of the coefficients of x^{32} and x^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The general term in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is given by $T_{r+1} = \binom{15}{r} (x^4)^{15-r} (-x^{-3})^r = \binom{

Vedclass · Accepted Answer

$1260$

Vedclass · Answer

(B) The general term in the expansion of $\left(x^4-\frac{1}{x^3}ight)^{15}$ is given by $T_{r+1} = \binom{15}{r} (x^4)^{15-r} (-x^{-3})^r = \binom{15}{r} (-1)^r x^{60-4r-3r} = \binom{15}{r} (-1)^r x^{60-7r}$.For the coefficient of $x^{32}$,we set $60-7r = 32$,which gives $7r = 28$,so $r = 4$.The coefficient is $\binom{15}{4} (-1)^4 = \frac{15 	imes 14 	imes 13 	imes 12}{4 	imes 3 	imes 2 	imes 1} = 1365$.For the coefficient of $x^{-31}$,we set $60-7r = -31$,which gives $7r = 91$,so $r = 13$.The coefficient is $\binom{15}{13} (-1)^{13} = \binom{15}{2} (-1) = -\frac{15 	imes 14}{2 	imes 1} = -105$.The sum of the coefficients is $1365 + (-105) = 1260$.

The sum of the coefficients of $x^{32}$ and $x^{-31}$ in the expansion of $\left(x^4-\frac{1}{x^3}\right)^{15}$ is:

Similar Questions

The term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ is:

The number of all possible values of $k$ for which the expansion $(\sqrt{x}+\sqrt[k]{y})^{10}$ will have exactly nine irrational terms is

The expansion of $(a+x)^n$ contains $15$ terms. When $x=1$,the ratio of the neighboring terms to the middle term in this expansion is $16$. Then the positive integral value of '$a$' is

If the coefficient of $x$ in the expansion of $\left(x^2+\frac{k}{x}\right)^5$ is $270$,then $k$ is equal to

The coefficient of $x^5$ in the expansion of $\left(2 x^3-\frac{1}{3 x^2}\right)^5$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module