Let K be the sum of the coefficients of the o | vedclass

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

Question

(C) The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is $K = 2^{99-1} = 2^{98}$.The middle term $a$ in the expansion of $(2+\frac{1}{\sqrt

Vedclass · Accepted Answer

$(50, 101)$

Vedclass · Answer

(C) The sum of coefficients of odd powers of $x$ in $(1+x)^{99}$ is $K = 2^{99-1} = 2^{98}$.The middle term $a$ in the expansion of $(2+\frac{1}{\sqrt{2}})^{200}$ is the $101^{st}$ term:$a = {}^{200}C_{100} (2)^{100} (\frac{1}{\sqrt{2}})^{100} = {}^{200}C_{100} \cdot 2^{100} \cdot 2^{-50} = {}^{200}C_{100} \cdot 2^{50}$.Now,consider the expression $\frac{{}^{200}C_{99} K}{a} = \frac{{}^{200}C_{99} \cdot 2^{98}}{{}^{200}C_{100} \cdot 2^{50}}$.Using the property ${}^{n}C_{r} = \frac{n-r+1}{r} {}^{n}C_{r-1}$,we have $\frac{{}^{200}C_{99}}{{}^{200}C_{100}} = \frac{100}{200-100+1} = \frac{100}{101}$.Thus,$\frac{{}^{200}C_{99} K}{a} = \frac{100}{101} \cdot 2^{98-50} = \frac{100}{101} \cdot 2^{48} = \frac{25 \cdot 4}{101} \cdot 2^{48} = \frac{25 \cdot 2^2 \cdot 2^{48}}{101} = \frac{2^{50} \cdot 25}{101}$.Comparing this with $\frac{2^{\ell} m}{n}$,we get $\ell = 50$,$m = 25$,and $n = 101$.Since $m=25$ and $n=101$ are odd,the ordered pair $(\ell, n)$ is $(50, 101)$.

Similar Questions

The number of rational terms in the binomial expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is

Let $n$ be a positive integer. If the coefficients of $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(1+x)^n$ are in $A$.$P$.,then the value of $n$ is:

The greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$ is

The coefficient of $x^{256}$ in the expansion of $(1-x)^{101}(x^{2}+x+1)^{100}$ is:

Prove that the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n}$ is twice the coefficient of $x^{n}$ in the expansion of $(1+x)^{2n-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module