If the coefficients of the (2r+6)^{text{th}} | vedclass

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

Question

(C) The general term in the expansion of $(1+x)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} x^{k}$.For the expansion of $(1+x)^{21}$,the coefficient of th

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The general term in the expansion of $(1+x)^{n}$ is given by $T_{k+1} = {}^{n}C_{k} x^{k}$.For the expansion of $(1+x)^{21}$,the coefficient of the $(k+1)^{	ext{th}}$ term is ${}^{21}C_{k}$.The coefficient of the $(2r+6)^{	ext{th}}$ term is ${}^{21}C_{(2r+6)-1} = {}^{21}C_{2r+5}$.The coefficient of the $(r-1)^{	ext{th}}$ term is ${}^{21}C_{(r-1)-1} = {}^{21}C_{r-2}$.Given that these coefficients are equal,we have ${}^{21}C_{2r+5} = {}^{21}C_{r-2}$.Using the property ${}^{n}C_{a} = {}^{n}C_{b}$,which implies either $a = b$ or $a + b = n$:Case $1$: $2r+5 = r-2 \Rightarrow r = -7$ (Not possible as $r$ must be a positive integer).Case $2$: $(2r+5) + (r-2) = 21$ $\Rightarrow 3r + 3 = 21$ $\Rightarrow 3r = 18$ $\Rightarrow r = 6$.

If the coefficients of the $(2r+6)^{\text{th}}$ and $(r-1)^{\text{th}}$ terms in the expansion of $(1+x)^{21}$ are equal,then the value of $r$ is:

Similar Questions

The term independent of $x(x>0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$ is:

The middle term in the expansion of $(1 + x)^{2n}$ is

In the expansion of ${\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9}$,the term independent of $x$ is

The numerically greatest term in the expansion of $(3x - 16y)^{15}$ when $x = \frac{2}{3}$ and $y = \frac{3}{2}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module