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

If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$, where $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is .... .

The sum of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$ is equal to

The coefficient of $x^4$ in the expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ is

Suppose $l, m, n$ respectively represent the coefficient of $x^{10}$,the constant term,and the coefficient of $x^{-10}$ in the expansion of $\left(a x^2+\frac{b}{x^3}\right)^{15}$. If $\frac{l}{m}+\frac{m}{n}=\frac{26}{11}$,then $a^2: b^2=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module