If the constant term in the binomial expansio | 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 $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (x^{1/2})^{10-r} (

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (x^{1/2})^{10-r} (-k x^{-2})^{r}$$T_{r+1} = {}^{10}C_{r} x^{(10-r)/2} (-k)^{r} x^{-2r}$$T_{r+1} = {}^{10}C_{r} (-k)^{r} x^{(10-5r)/2}$For the constant term,the exponent of $x$ must be $0$:$\frac{10-5r}{2} = 0$ $\Rightarrow 5r = 10$ $\Rightarrow r = 2$Substituting $r = 2$ into the expression for $T_{r+1}$:$T_{3} = {}^{10}C_{2} (-k)^{2} = 405$$45 \cdot k^{2} = 405$$k^{2} = \frac{405}{45} = 9$$|k| = \sqrt{9} = 3$

If the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is $405$,then $|k|$ equals

Similar Questions

The number of integral terms in the expansion of $(5^{\frac{1}{2}} + 7^{\frac{1}{8}})^{1016}$ is

The coefficients of three consecutive terms of $(1+x)^{n+5}$ are in the ratio $5: 10: 14$. Then $n=$

Find $a$ if the $17^{\text{th}}$ and $18^{\text{th}}$ terms of the expansion $(2 + a)^{50}$ are equal.

If the sum of the coefficients in the expansion of $(x + y)^n$ is $1024$,then the value of the greatest coefficient in the expansion is

If the $7^{th}$ term from the beginning in the binomial expansion of ${\left( {\frac{3}{{{{\left( {84} \right)}^{\frac{1}{3}}}}} + \sqrt 3 \ln x} \right)^9}$ for $x > 0$ is equal to $729$,then the possible value of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module