Given that the term of the expansion (x^{1/3} | vedclass

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

Question

(C) The general term $T_{r + 1}$ in the expansion of $(x^{1/3} - x^{-1/2})^{15}$ is given by:$T_{r + 1} = ^{15}C_r (x^{1/3})^{15 - r} (-x^{-1/2})^r$$T

Vedclass · Accepted Answer

$1001$

Vedclass · Answer

(C) The general term $T_{r + 1}$ in the expansion of $(x^{1/3} - x^{-1/2})^{15}$ is given by:$T_{r + 1} = ^{15}C_r (x^{1/3})^{15 - r} (-x^{-1/2})^r$$T_{r + 1} = ^{15}C_r (-1)^r x^{(15 - r)/3} x^{-r/2}$$T_{r + 1} = ^{15}C_r (-1)^r x^{(15 - r)/3 - r/2}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$\frac{15 - r}{3} - \frac{r}{2} = 0$$2(15 - r) - 3r = 0$$30 - 2r - 3r = 0$$5r = 30 \Rightarrow r = 6$Substituting $r = 6$ into the general term:$T_7 = ^{15}C_6 (-1)^6 = ^{15}C_6 = \frac{15 	imes 14 	imes 13 	imes 12 	imes 11 	imes 10}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1} = 5005$Given $T_7 = 5m$,we have $5005 = 5m \Rightarrow m = 1001$.

Given that the term of the expansion $(x^{1/3} - x^{-1/2})^{15}$ which does not contain $x$ is $5m$,where $m \in N$,then $m =$

Similar Questions

The coefficient of $x^{7}$ in the expression $(1+x)^{10}+x(1+x)^{9}+x^{2}(1+x)^{8}+\ldots+x^{10}$ is

The coefficients of two consecutive terms in the expansion of $(1 + x)^n$ will be equal,if

If $21^{\text{st}}$ and $22^{\text{nd}}$ terms in the expansion of $(1+x)^{44}$ are equal,then $x$ is equal to

If the coefficients of $x^{-2}$ and $x^{-4}$ in the expansion of ${\left( {{x^{\frac{1}{3}}} + \frac{1}{{2{x^{\frac{1}{3}}}}}} \right)^{18}}, (x > 0),$ are $m$ and $n$ respectively,then $\frac{m}{n}$ is equal to

If the fourth term in the binomial expansion of $\left(\sqrt{\frac{1}{x^{1+\log _{10} x}}}+x^{\frac{1}{12}}\right)^{6}$ is equal to $200$,and $x > 1$,then the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module