If the coefficient of x^10 in the binomial expansion of ( x5^1/4+√5x^1/3)^60 is 5^k l, whe…

Question

If the coefficient of $x^{10}$ in the binomial expansion of $\left(\dfrac{\sqrt x}{5^{1/4}}+\dfrac{\sqrt5}{x^{1/3}}\right)^{60}$ is $5^k\cdot l$, where $l,k\in N$ and $l$ is coprime to 5, then $k$ is equal to

Answer: 5

Accepted Answer

$$T_{r+1}=\binom{60}{r}\left(5^{-1/4}\right)^{60-r}x^{(60-r)/2}\left(5^{1/2}\right)^r x^{-r/3}=\binom{60}{r}5^{\frac{3r-60}{4}}\,x^{\frac{180-5r}{6}}$$

For $x^{10}$: $\dfrac{180-5r}{6}=10\Rightarrow r=24$.

Coefficient $=\binom{60}{24}\cdot5^{\frac{3(24)-60}{4}}=\binom{60}{24}\cdot5^3=\dfrac{60!}{24!\,36!}\times5^3$

Exponent of 5 in $60!$: $12+2=14$. In $24!$: $4$. In $36!$: $7+1=8$.

Net exponent of 5 in $\dfrac{60!}{24!\,36!}$: $14-4-8=2$; adding the extra $5^3$: total exponent $=2+3=5$.

$k=5$

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