In the expansion of {left( x - frac{1}{x} rig | vedclass

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

Question

(A) The general term in the expansion of ${\left( x - \frac{1}{x} \right)^6}$ is given by $T_{r+1} = {^6C_r} \cdot x^{6-r} \cdot \left( -\frac{1}{x} \

Vedclass · Accepted Answer

$-20$

Vedclass · Answer

(A) The general term in the expansion of ${\left( x - \frac{1}{x} \right)^6}$ is given by $T_{r+1} = {^6C_r} \cdot x^{6-r} \cdot \left( -\frac{1}{x} \right)^r$.Simplifying this,we get $T_{r+1} = {^6C_r} \cdot (-1)^r \cdot x^{6-r} \cdot x^{-r} = {^6C_r} \cdot (-1)^r \cdot x^{6-2r}$.For the term to be independent of $x$ (the constant term),the exponent of $x$ must be $0$,so $6 - 2r = 0$,which implies $r = 3$.Substituting $r = 3$ into the expression,the constant term is ${^6C_3} \cdot (-1)^3 = 20 \cdot (-1) = -20$.

In the expansion of ${\left( x - \frac{1}{x} \right)^6}$,the constant term is

Similar Questions

If the $17^{\text{th}}$ and the $18^{\text{th}}$ terms in the expansion of $(2+a)^{50}$ are equal,then the coefficient of $x^{35}$ in the expansion of $(a+x)^{-2}$ is

The second,third and fourth terms in the binomial expansion $(x+a)^n$ are $240, 720$ and $1080$ respectively. Find $x, a$ and $n$.

Write the general term in the expansion of $(x^{2}-yx)^{12}, x \neq 0$.

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$,$n \in N$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

If the $r$-th and $(r+1)$-th terms in the expansion of $(p+q)^{n}$ are equal,then the value of $\frac{(n+1)q}{r(p+q)}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module