p, q are two prime numbers. For n=pq,if the e | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $\left(x^{-5/4} + 2x^{4/5}\right)^n$ is given by:$T_{r+1} = {}^nC_r (x^{-5/4})^{n-r} (2x^{4/5})^r =

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$82$

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(A) The general term $T_{r+1}$ in the expansion of $\left(x^{-5/4} + 2x^{4/5}ight)^n$ is given by:$T_{r+1} = {}^nC_r (x^{-5/4})^{n-r} (2x^{4/5})^r = {}^nC_r 2^r x^{-\frac{5n}{4} + \frac{5r}{4} + \frac{4r}{5}} = {}^nC_r 2^r x^{\frac{-25n + 41r}{20}}$.For the coefficient of $x^0$ to be non-zero,the exponent must be zero:$-25n + 41r = 0 \Rightarrow r = \frac{25n}{41}$.Since $r$ is an integer and $0 \le r \le n$,$n$ must be a multiple of $41$. Given $n = pq$ where $p, q$ are primes,$n$ could be $41 	imes 1 = 41$ (not possible as $1$ is not prime) or $41 	imes 2 = 82$.For the coefficient of $x^{-n}$ to be non-zero:$\frac{-25n + 41r}{20} = -n$ $\Rightarrow -25n + 41r = -20n$ $\Rightarrow 41r = 5n$ $\Rightarrow r = \frac{5n}{41}$.For $r$ to be an integer,$n$ must be a multiple of $41$. The smallest such $n = pq$ is $41 	imes 2 = 82$.

$p, q$ are two prime numbers. For $n=pq$,if the expansion $\left(x^{-5/4} + 2x^{4/5}\right)^n$ contains non-zero coefficients of $x^{-n}$ and $x^0$,then the least value of such $n$ is

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