The number of rational terms in the binomial | 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 $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is given by $T_{r+1} = {}^{100}C_r (5^{1/4})^{100-r} (4^{1/5})^r$. This sim

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The general term in the expansion of $(\sqrt[4]{5}+\sqrt[5]{4})^{100}$ is given by $T_{r+1} = {}^{100}C_r (5^{1/4})^{100-r} (4^{1/5})^r$. This simplifies to $T_{r+1} = {}^{100}C_r (5)^{\frac{100-r}{4}} (2^2)^{\frac{r}{5}} = {}^{100}C_r (5)^{\frac{100-r}{4}} (2)^{\frac{2r}{5}}$. For the term to be rational,the exponents of $5$ and $2$ must be integers. For $\frac{100-r}{4}$ to be an integer,$r$ must be a multiple of $4$. Since $0 \le r \le 100$,$r \in \{0, 4, 8, \dots, 100\}$. For $\frac{2r}{5}$ to be an integer,$r$ must be a multiple of $5$. Since $0 \le r \le 100$,$r \in \{0, 5, 10, \dots, 100\}$. Thus,$r$ must be a multiple of both $4$ and $5$,which means $r$ must be a multiple of $	ext{lcm}(4, 5) = 20$. The possible values for $r$ are $0, 20, 40, 60, 80, 100$. There are $6$ such values,so the number of rational terms is $6$.

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

Similar Questions

Find the coefficient of $x^{6} y^{3}$ in the expansion of $(x+2 y)^{9}$.

The $13^{th}$ term in the expansion of $\left(x^{2}+\frac{2}{x}\right)^{n}$ is independent of $x$. Then the sum of the divisors of $n$ is:

If the coefficients of the $r$-th,$(r+1)$-th,and $(r+2)$-th terms in the expansion of $(1+x)^n$ are respectively in the ratio $2:4:5$,then $(r, n) =$

The largest term in the expansion of $(3 + 2x)^{50}$ where $x = \frac{1}{5}$ is

The coefficient of $x^n$ in the expansion of $(1 - 2x + 3x^2 - 4x^3 + \dots)^{-n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module