The ratio of the coefficient of the middle te | vedclass

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

Question

(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.Its coefficient is $^{20}C_{10}$.The m

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The middle term in the expansion of $(1+x)^{20}$ is the $\left(\frac{20}{2} + 1\right)^{th} = 11^{th}$ term.Its coefficient is $^{20}C_{10}$.The middle terms in the expansion of $(1+x)^{19}$ are the $\left(\frac{19+1}{2}\right)^{th} = 10^{th}$ and $\left(\frac{19+1}{2} + 1\right)^{th} = 11^{th}$ terms.Their coefficients are $^{19}C_{9}$ and $^{19}C_{10}$.The sum of these coefficients is $^{19}C_{9} + ^{19}C_{10}$.Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,we have $^{19}C_{9} + ^{19}C_{10} = ^{20}C_{10}$.Therefore,the required ratio is $\frac{^{20}C_{10}}{^{19}C_{9} + ^{19}C_{10}} = \frac{^{20}C_{10}}{^{20}C_{10}} = 1$.

The ratio of the coefficient of the middle term in the expansion of $(1+x)^{20}$ and the sum of the coefficients of two middle terms in the expansion of $(1+x)^{19}$ is $....$

Similar Questions

If the $m$-th term is the middle term in the expansion of $\left(x^2-\frac{1}{2x}\right)^{20}$,find the coefficient of $T_{m+3}$.

The coefficient of $t^{50}$ in $(1 + t^2)^{25}(1 + t^{25})(1 + t^{40})(1 + t^{45})(1 + t^{47})$ is

If $x + y = 1$,then $\sum\limits_{r = 0}^n {{r^2}{\,^n}{C_r}{x^r}{y^{n - r}}} $ equals

The number of terms in the expansion of ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$ which are integers is

The number of irrational terms in the expansion of $(5^{1/2} + 7^{1/8})^{1024} + (5^{1/2} - 7^{1/8})^{1024}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module