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

(C) Simplify the expression inside the bracket:$\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-

Vedclass · Accepted Answer

$210$

Vedclass · Answer

(C) Simplify the expression inside the bracket:$\frac{x+1}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3})^3+1^3}{x^{2/3}-x^{1/3}+1} = \frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$$\frac{x-1}{x-x^{1/2}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)} = \frac{\sqrt{x}+1}{\sqrt{x}} = 1+x^{-1/2}$Now,the expression becomes: $(x^{1/3}+1 - (1+x^{-1/2}))^{10} = (x^{1/3}-x^{-1/2})^{10}$The general term $T_{r+1}$ is given by: $^{10}C_r (x^{1/3})^{10-r} (-x^{-1/2})^r = ^{10}C_r (-1)^r x^{\frac{10-r}{3} - \frac{r}{2}}$For the term to be independent of $x$,the exponent must be $0$:$\frac{10-r}{3} - \frac{r}{2} = 0$$20-2r-3r = 0$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$Substituting $r=4$ into the coefficient part:$T_{5} = ^{10}C_4 = \frac{10 	imes 9 	imes 8 	imes 7}{4 	imes 3 	imes 2 	imes 1} = 210$

The term independent of $x$ in the expansion of ${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ is

Similar Questions

The sum of all rational terms in the expansion of $(2+\sqrt{3})^8$ is

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

The $r^{th}$ term in the expansion of $(a + 2x)^n$ is

Let $m$ be the smallest positive integer such that the coefficient of $x^2$ in the expansion of $(1+x)^2+(1+x)^3+\cdots+(1+x)^{49}+(1+mx)^{50}$ is $(3n+1)^{51}C_3$ for some positive integer $n$. Then the value of $n$ is

If the coefficients of $x^3$ and $x^4$ in the expansion of $(1 + ax + bx^2)(1 - 2x)^{18}$ in powers of $x$ are both zero,then $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module