Question

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

Answer 1

Answer: 5

$T_{r+1}={}^{60}{C}_r{\left(x^{\frac{1}{2}}\right)}^{60-r}{\left(x^{-\frac{1}{3}}\right)}^r{\left(5^{\frac{-1}{4}}\right)}^{60-r}{\left(\frac{1}{5^2}\right)}^r$
$\text{ For }{x}^{10}\frac{60-r}{2}-\frac{r}{3}=10$
$\Rightarrow 180-3r-2r=60$
$\Rightarrow r=24$
$\therefore \text{ Coeff. of }{x}^{10}=\frac{{}^{60}{C}_{24}}{5^9}{5}^{12}=5^{k}I$
$\text{ as I and }5\text{ are coprime }$
$k=3+\text{ exponent of }5\text{ in }{}^{60}{C}_{24}$
$=3+\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]-\left[\frac{24}{5}\right]-\left[\frac{24}{5^2}\right]-\left[\frac{36}{5}\right]-\left[\frac{36}{5^2}\right]$
$=3+(12+2-4-0-7-1)$
$=3+2=5$