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 coefficient of ${x^{39}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

If for some positive integer $n,$ the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+5}$ are in the ratio $5: 10: 14,$ then the largest coefficient in this expansion is

If $\left(\frac{3^{6}}{4^{4}}\right) k$ is the term independent of $x$ in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$,then $k$ is equal to ...... .

The term independent of $x$ in the expansion of $\left(\frac{x+1}{x^{2/3}+1-x^{1/3}}-\frac{x-1}{x-x^{1/2}}\right)^{10}, x>1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module