The middle term in the expansion of {left( {x | vedclass

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

Question

(B) For the expansion of $(x + a)^{2n}$,the total number of terms is $2n + 1$,which is odd. Therefore,the middle term is the $(n+1)$-th term.Using the

Vedclass · Accepted Answer

$\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n - 1)}{n!}$

Vedclass · Answer

(B) For the expansion of $(x + a)^{2n}$,the total number of terms is $2n + 1$,which is odd. Therefore,the middle term is the $(n+1)$-th term.Using the general term formula $T_{r+1} = {}^{2n}C_r (x)^{2n-r} (\frac{1}{2x})^r$,for the middle term,we set $r = n$.$T_{n+1} = {}^{2n}C_n (x)^{2n-n} (\frac{1}{2x})^n = {}^{2n}C_n \cdot x^n \cdot \frac{1}{2^n x^n} = \frac{(2n)!}{n! n! 2^n}$.Expanding $(2n)! = 1 \cdot 2 \cdot 3 \cdot \dots \cdot (2n)$,we separate the odd and even terms:$(2n)! = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdot \dots \cdot 2n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot [1 \cdot 2 \cdot 3 \cdot \dots \cdot n] = [1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!$.Substituting this back into the expression:$T_{n+1} = \frac{[1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)] \cdot 2^n \cdot n!}{n! \cdot n! \cdot 2^n} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{n!}$.

The middle term in the expansion of ${\left( {x + \frac{1}{{2x}}} \right)^{2n}}$ is:

Similar Questions

If $\frac{T_2}{T_3}$ in the expansion of $(a + b)^n$ and $\frac{T_3}{T_4}$ in the expansion of $(a + b)^{n + 3}$ are equal,then $n=$

If $(1+x)^{15}=a_0+a_1 x+\ldots+a_{15} x^{15}$,then $\sum_{r=1}^{15} r \frac{a_r}{a_{r-1}}$ is equal to

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$,is $\alpha \times 2^8 \times \sqrt[5]{3}$,then $25 \alpha$ is equal to :

$x^r$ occurs in the expansion of ${\left( x^3 + \frac{1}{x^4} \right)^n}$ provided -

The greatest term in the expansion of $(1+x)^{15}$,when $x=\frac{1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module