If the coefficient of x^{10} in the binomial | vedclass

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(A) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{1/2}}{5^{1/4}} + \frac{5^{1/2}}{x^{1/3}}\right)^{60}$ is given by:$T_{r+1} = {}^{60

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$5$

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(A) The general term $T_{r+1}$ in the expansion of $\left(\frac{x^{1/2}}{5^{1/4}} + \frac{5^{1/2}}{x^{1/3}}\right)^{60}$ is given by:$T_{r+1} = {}^{60}C_r \left(x^{1/2} \cdot 5^{-1/4}\right)^{60-r} \left(5^{1/2} \cdot x^{-1/3}\right)^r$$T_{r+1} = {}^{60}C_r \cdot x^{(60-r)/2} \cdot 5^{-(60-r)/4} \cdot 5^{r/2} \cdot x^{-r/3}$$T_{r+1} = {}^{60}C_r \cdot x^{(60-r)/2 - r/3} \cdot 5^{r/2 - (60-r)/4}$For the coefficient of $x^{10}$,we set the exponent of $x$ to $10$:$\frac{60-r}{2} - \frac{r}{3} = 10$ $\Rightarrow \frac{180-3r-2r}{6} = 10$ $\Rightarrow 180-5r = 60$ $\Rightarrow 5r = 120$ $\Rightarrow r = 24$Now,substitute $r=24$ into the coefficient part:Coefficient $= {}^{60}C_{24} \cdot 5^{24/2 - (60-24)/4} = {}^{60}C_{24} \cdot 5^{12 - 9} = {}^{60}C_{24} \cdot 5^3$We need to find the power of $5$ in ${}^{60}C_{24} = \frac{60!}{24!36!}$.Using Legendre's formula,the exponent of $5$ in $n!$ is $E_5(n!) = \lfloor n/5 \rfloor + \lfloor n/25 \rfloor$.$E_5(60!) = 12 + 2 = 14$$E_5(24!) = 4 + 0 = 4$$E_5(36!) = 7 + 1 = 8$Exponent of $5$ in ${}^{60}C_{24} = 14 - (4+8) = 2$.Thus,the total power of $5$ is $2 + 3 = 5$.Therefore,$k = 5$.

If the coefficient of $x^{10}$ in the binomial expansion of $\left(\frac{\sqrt{x}}{5^{1/4}}+\frac{\sqrt{5}}{x^{1/3}}\right)^{60}$ is $5^k l$,where $l, k \in N$ and $l$ is coprime to $5$,then $k$ is equal to

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