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$$T_{r+1} =

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$$T_{r+1} = {}^{100}C_r 5^{\frac{100-r}{4}} 4^{\frac{r}{5}}$For the term to be rational,the exponents of $5$ and $4$ must be integers.Thus,$\frac{100-r}{4}$ must be an integer,which implies $r$ must be a multiple of $4$ (i.e.,$r \in \{0, 4, 8, \dots, 100\}$).Also,$\frac{r}{5}$ must be an integer,which implies $r$ must be a multiple of $5$ (i.e.,$r \in \{0, 5, 10, \dots, 100\}$).Therefore,$r$ must be a multiple of $	ext{lcm}(4, 5) = 20$.The possible values for $r$ are $0, 20, 40, 60, 80, 100$.Counting these values,we get $6$ terms.Hence,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

The term independent of $x$ in the expansion of $\left[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}\right]^{10}, x \neq 1,$ is equal to ....... .

The coefficient of the term independent of $x$ in the expansion of $(1 + x + 2x^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is

The coefficient of $\frac{1}{x}$ in the expansion of $(1 + x)^n (1 + \frac{1}{x})^n$ is

Find the term independent of $x$ in the expansion of $\left(\sqrt[3]{x}+\frac{1}{2 \sqrt[3]{x}}\right)^{18}, x > 0$.

If $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$,then $2 a_2+3 a_3+4 a_4+\ldots+20 a_{20}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module