The number of terms in the expansion of {left | vedclass

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(B) The given expression is ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$.Simplifying the terms: $\sqrt[4]{9} = (3^2)^{1/4} = 3^{1/2}$ and $\sq

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$251$

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(B) The given expression is ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$.Simplifying the terms: $\sqrt[4]{9} = (3^2)^{1/4} = 3^{1/2}$ and $\sqrt[6]{8} = (2^3)^{1/6} = 2^{1/2}$.So,the expression is $(3^{1/2} + 2^{1/2})^{500}$.The general term in the expansion is $T_{r+1} = \binom{500}{r} (3^{1/2})^{500-r} (2^{1/2})^r = \binom{500}{r} 3^{(500-r)/2} 2^{r/2}$.For the term to be an integer,the exponents of $3$ and $2$ must be non-negative integers.This requires $(500-r)/2$ and $r/2$ to be integers.This implies $r$ must be an even integer such that $0 \le r \le 500$.The possible values for $r$ are $0, 2, 4, \dots, 500$.This is an arithmetic progression with first term $a = 0$,last term $l = 500$,and common difference $d = 2$.The number of terms is given by $n = \frac{l-a}{d} + 1 = \frac{500-0}{2} + 1 = 250 + 1 = 251$.

The number of terms in the expansion of ${\left( {\sqrt[4]{9} + \sqrt[6]{8}} \right)^{500}}$ which are integers is

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