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,$n = 10$,which is even.Therefore,the middl

Vedclass · Accepted Answer

$^{10}C_5$

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,$n = 10$,which is even.Therefore,the middle term is the $(\frac{10}{2} + 1) = 6$-th term,denoted as $T_6$.The general term $T_{r+1}$ in the expansion of $(x + a)^n$ is given by $T_{r+1} = ^nC_r x^{n-r} a^r$.For $T_6$,we have $r = 5$.Substituting the values,$T_6 = ^{10}C_5 (x)^{10-5} (\frac{1}{x})^5$.$T_6 = ^{10}C_5 x^5 \cdot \frac{1}{x^5} = ^{10}C_5$.Thus,the middle term is $^{10}C_5$.

The middle term in the expansion of ${\left( x + \frac{1}{x} \right)^{10}}$ is

Similar Questions

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

The middle term in the expansion of $(1 + 3x + 3x^2 + x^3)^6$ is

The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is independent of $x$. Then the sum of the divisors of $n$ is:

In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

If $a^3 + b^6 = 2$,then the maximum value of the term independent of $x$ in the expansion of $(ax^{1/3} + bx^{-1/6})^9$ is,where $(a > 0, b > 0)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module