The least value of n for which the number of | vedclass

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(A) The general term of the expansion is $T_{r+1} = {}^{n}C_{r} (7^{1/3})^{n-r} (11^{1/12})^{r} = {}^{n}C_{r} 7^{(n-r)/3} 11^{r/12}$.For the term to b

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$2184$

Vedclass · Answer

(A) The general term of the expansion is $T_{r+1} = {}^{n}C_{r} (7^{1/3})^{n-r} (11^{1/12})^{r} = {}^{n}C_{r} 7^{(n-r)/3} 11^{r/12}$.For the term to be an integer,both exponents $\frac{n-r}{3}$ and $\frac{r}{12}$ must be integers.This implies $r$ must be a multiple of $12$,so $r \in \{0, 12, 24, \dots, 12k\}$.Also,$n-r$ must be a multiple of $3$. Since $r$ is a multiple of $12$,$r$ is also a multiple of $3$,which implies $n$ must be a multiple of $3$.There are $183$ integral terms,so $r$ takes values $0, 12, 24, \dots, 12 	imes 182$.The maximum value of $r$ is $12 	imes 182 = 2184$.Since $r \le n$,the smallest $n$ that allows $183$ terms (where $r$ goes up to $2184$) is $n = 2184$.

The least value of $n$ for which the number of integral terms in the Binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^{n}$ is $183$ is:

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