Coefficient of middle term in expansion of {(1 + x)^{10}} is

English

Gujarati

Hindi

7.Binomial Theorem

easy

The coefficient of middle term in the expansion of ${(1 + x)^{10}}$ is

A

$\frac{{10!}}{{5!\,6!}}$

B

$\frac{{10\,!}}{{{{(5\,!)}^2}}}$

C

$\frac{{10\,!}}{{5\,!\,7\,!}}$

D

None of these

Solution

(b) Here, index $= 10$, an even number, therefore, middle term is ${\left( {\frac{{10 + 2}}{2}} \right)^{th}}$ i.e. $6^{th}$ term.

Also ,${T_6}\,\, = \,\,{}^{10}{C_5}\,{x^5}$

==>Coefficient of middle term $ = {}^{10}{C_5}.$

Standard 11

Mathematics

Share

Similar Questions

If the number of integral terms in the expansion of $\left(3^{\frac{1}{2}}+5^{\frac{1}{8}}\right)^{\text {n }}$ is exactly $33,$ then the least value of $n$ is

medium

(JEE MAIN-2020)

The sum of the real values of $x$ for which the middle term in the binomial expansion of ${\left( {\frac{{{x^3}}}{3} + \frac{3}{x}} \right)^8}$ equals $5670$ is

hard

(JEE MAIN-2019)

In the expansion of ${\left( {\frac{a}{x} + bx} \right)^{12}}$,the coefficient of $x^{-10}$ will be

medium

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is equal to

hard

(JEE MAIN-2023)

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

easy