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 $5^{k} l$, where $l, k \in \mathbb{N}$ and $l$ is coprime to 5, then k is equal to ______.

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Binomial Theorem

If the coefficient of

x^{10}

in the binomial expansion of

(\frac{x}{5 ^{\frac{1}{4}}} + \frac{5}{x ^{\frac{1}{3}}})^{60}

5^{k} l

, where

l, k \in N

and

l

is coprime to 5, then k is equal to ______.

Answer: 5

In the binomial expansion

(\frac{x ^{1/2}}{5 ^{1/4}} + \frac{5 ^{1/2}}{x ^{1/3}})^{60}

, the general term is:

T_{r + 1} =^{60} C_{r} (\frac{x ^{1/2}}{5 ^{1/4}})^{60 - r} (\frac{5 ^{1/2}}{x ^{1/3}})^{r} =^{60} C_{r} \cdot 5^{\frac{3 r - 60}{4}} \cdot x^{\frac{180 - 5 r}{6}}

For the coefficient of

x^{10}

, we need:

\frac{180 - 5 r}{6} = 10 \Rightarrow r = 24

The coefficient is:

^{60} C_{24} \cdot 5^{\frac{3 ( 24 ) - 60}{4}} =^{60} C_{24} \cdot 5^{3}

Now count the powers of 5 in

^{60} C_{24} = \frac{60 !}{24 ! \cdot 36 !}

Total power of 5:

14 - 4 - 8 + 3 = 5

Therefore,

k = 5

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Integer answer