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 $(x + a)^n$,if $n$ is even,the middle term is the $(\frac{n}{2} + 1)^{th}$ term.Here,$n = 18$,which is even.Therefore,the mid

Vedclass · Accepted Answer

$-^{18}C_9$

Vedclass · Answer

(B) For the expansion of $(x + a)^n$,if $n$ is even,the middle term is the $(\frac{n}{2} + 1)^{th}$ term.Here,$n = 18$,which is even.Therefore,the middle term is the $(\frac{18}{2} + 1)^{th} = 10^{th}$ term.The general term $T_{r+1}$ in the expansion of $(x - \frac{1}{x})^{18}$ is given by $T_{r+1} = ^{18}C_r (x)^{18-r} (- \frac{1}{x})^r$.For the $10^{th}$ term,$r = 9$.$T_{10} = ^{18}C_9 (x)^{18-9} (- \frac{1}{x})^9$$T_{10} = ^{18}C_9 (x)^9 (-1)^9 (\frac{1}{x})^9$$T_{10} = ^{18}C_9 (x)^9 (-1) (\frac{1}{x^9})$$T_{10} = -^{18}C_9$.

The middle term in the expansion of ${\left( {x - \frac{1}{x}} \right)^{18}}$ 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 $n=$

The coefficient of $t^{12}$ in the expansion of $(1 + t^2)^6(1 + t^6)(1 + t^{12})$ is:

What is the coefficient of $x^{100}$ in $(1 + x + x^2 + x^3 + \dots + x^{100})^3$?

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 coefficients of $x^2$ and $x^3$ in the expansion of $(3 + ax)^9$ are the same,then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module