Find the middle term of the expansion of lef | vedclass

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

Question

(B) The given expression is $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $. Here,the power $ n = 10 $,which is an even number. Therefore,the number o

Vedclass · Accepted Answer

$ {}^{10}C_{5} $

Vedclass · Answer

(B) The given expression is $ \left(\frac{10}{x}+\frac{x}{10}ight)^{10} $. Here,the power $ n = 10 $,which is an even number. Therefore,the number of terms in the expansion is $ n + 1 = 11 $. The middle term is the $ \left(\frac{n}{2} + 1ight) $-th term,which is $ \left(\frac{10}{2} + 1ight) = 6 $-th term. The general term of the expansion $ (a+b)^n $ is given by $ T_{r+1} = {}^{n}C_{r} a^{n-r} b^{r} $. For the $ 6 $-th term,$ r = 5 $. $ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}ight)^{10-5} \left(\frac{x}{10}ight)^{5} $. $ T_{6} = {}^{10}C_{5} \left(\frac{10}{x}ight)^{5} \left(\frac{x}{10}ight)^{5} $. $ T_{6} = {}^{10}C_{5} 	imes 1 = {}^{10}C_{5} $.

Find the middle term of the expansion of $ \left(\frac{10}{x}+\frac{x}{10}\right)^{10} $.

Similar Questions

The binomial coefficients which are in decreasing order are

Let the coefficients of $x^{-1}$ and $x^{-3}$ in the expansion of $(2x^{1/5} - x^{-1/5})^{15}$,$x > 0$,be $m$ and $n$ respectively. If $r$ is a positive integer such that $mn^2 = {}^{15}C_r \cdot 2^r$,then the value of $r$ is equal to

In the expansion of $(\sqrt[5]{3}+\sqrt[3]{2})^{15}$

If the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $(\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}})^n$ is $\sqrt{6} : 1$,then the third term from the beginning is:

Find the term independent of $x$ in the expansion of $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{6}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module