The sum of all those terms which are rational | vedclass

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

Question

(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le

Vedclass · Accepted Answer

$43$

Vedclass · Answer

(D) The general term of the expansion $(2^{1/3} + 3^{1/4})^{12}$ is given by $T_{r+1} = ^{12}C_{r} (2^{1/3})^{12-r} (3^{1/4})^{r}$,where $0 \le r \le 12$.For the term to be rational,the exponents of $2$ and $3$ must be integers.This implies $\frac{12-r}{3}$ must be an integer,so $r$ must be a multiple of $3$ $(r \in \{0, 3, 6, 9, 12\})$.Also,$\frac{r}{4}$ must be an integer,so $r$ must be a multiple of $4$ $(r \in \{0, 4, 8, 12\})$.The common values for $r$ are $r = 0$ and $r = 12$.For $r = 0$: $T_{1} = ^{12}C_{0} (2^{1/3})^{12} (3^{1/4})^{0} = 1 	imes 2^{4} 	imes 1 = 16$.For $r = 12$: $T_{13} = ^{12}C_{12} (2^{1/3})^{0} (3^{1/4})^{12} = 1 	imes 1 	imes 3^{3} = 27$.The sum of these rational terms is $16 + 27 = 43$.

The sum of all those terms which are rational numbers in the expansion of $(2^{1/3} + 3^{1/4})^{12}$ is:

Similar Questions

If the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end in the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^n$ is $\frac{1}{6}$,then $n=$

The constant term in the expansion of $\left(x^{2}-\frac{1}{x^{2}}\right)^{16}$ is

In the expansion of ${\left( \frac{1 + x}{1 - x} \right)^2}$,the coefficient of ${x^n}$ is:

If the term independent of $x$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^2}\right)^{10}$ is $405$,then $k=$

If coefficients of $2^{nd}$,$3^{rd}$ and $4^{th}$ terms in the binomial expansion of $(1 + x)^n$ are in $A.P.$,then $n^2 - 9n$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module