The term independent of x in the expansion of | vedclass

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

Question

(A) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is given by: $T_{r+1} = { }^{18} C_r (\sqrt{x})^{18

Vedclass · Accepted Answer

$-{ }^{18} C_9 2^9$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is given by: $T_{r+1} = { }^{18} C_r (\sqrt{x})^{18-r} \left(-\frac{2}{\sqrt{x}}\right)^r$ $T_{r+1} = { }^{18} C_r (x)^{\frac{18-r}{2}} (-2)^r (x)^{-\frac{r}{2}}$ $T_{r+1} = { }^{18} C_r (-2)^r x^{\frac{18-r-r}{2}}$ $T_{r+1} = { }^{18} C_r (-2)^r x^{9-r}$ For the term to be independent of $x$,the exponent of $x$ must be $0$: $9 - r = 0 \Rightarrow r = 9$ Substituting $r = 9$ into the general term: $T_{9+1} = { }^{18} C_9 (-2)^9$ $T_{10} = -{ }^{18} C_9 2^9$

The term independent of $x$ in the expansion of $\left(\sqrt{x}-\frac{2}{\sqrt{x}}\right)^{18}$ is

Similar Questions

The numerically greatest term in the expansion of $(2x - 3y)^{13}$ when $x = \frac{7}{2}$ and $y = \frac{3}{7}$ is:

If the $6^{th}$ term in $\left(\frac{2p}{3} + \frac{3q}{2}\right)^9$ is $ap^bq^c$,then $a, b$ and $c$ respectively are

If the $m$-th term is the middle term in the expansion of $\left(x^2-\frac{1}{2x}\right)^{20}$,find the coefficient of $T_{m+3}$.

In the binomial expansion of $(a - b)^n, n \ge 5,$ the sum of the $5^{th}$ and $6^{th}$ terms is zero. Then $a/b$ equals:

The numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module