Understanding the Binomial Expansion and General Term

The problem asks for the coefficient of $x^{10}$ in the binomial expansion of $\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt 5 } \over {{x^{{1 \over 3}}}}}} \right)^{60}$. To find the coefficient of a specific term in a binomial expansion of the form $(A+B)^n$, we utilize the general term formula: $T_{r+1} = {}^{n}C_r A^{n-r} B^r$ First, let's identify the components $A$, $B$, and $n$ from the given expression and rewrite them in a more convenient form with powers: $A = \frac{\sqrt{x}}{5^{1/4}} = \frac{x^{1/2}}{5^{1/4}} = 5^{-1/4} x^{1/2}$ $B = \frac{\sqrt{5}}{x^{1/3}} = \frac{5^{1/2}}{x^{1/3}} = 5^{1/2} x^{-1/3}$ $n = 60$

Step 1: Write down the General Term ($T_{r+1}$)

Substitute the identified expressions for $A$, $B$, and $n$ into the general term formula. This step sets up the expression that we will manipulate to find the desired coefficient. $T_{r+1} = {}^{60}C_r \left( 5^{-1/4} x^{1/2} \right)^{60-r} \left( 5^{1/2} x^{-1/3} \right)^r$

Step 2: Simplify the powers of $x$ and $5$

To determine the coefficient of $x^{10}$, we need to isolate the powers of $x$ and $5$ within the general term. We achieve this by applying exponent rules $(a^m)^n = a^{mn}$ and $a^m \cdot a^n = a^{m+n}$.

First, distribute the outer exponents to the terms inside the parentheses: $T_{r+1} = {}^{60}C_r \cdot 5^{(-1/4)(60-r)} \cdot x^{(1/2)(60-r)} \cdot 5^{(1/2)r} \cdot x^{(-1/3)r}$ $T_{r+1} = {}^{60}C_r \cdot 5^{(-15 + r/4)} \cdot x^{(30 - r/2)} \cdot 5^{r/2} \cdot x^{-r/3}$

Next, group the terms with the same base (i.e., $x$ and $5$) and combine their exponents:

Exponent of $x$: We combine the exponents of $x$: $(30 - r/2) + (-r/3)$. To add these fractions, we find a common denominator, which is $6$: $30 - \frac{r}{2} - \frac{r}{3} = 30 - \frac{3r}{6} - \frac{2r}{6} = 30 - \frac{5r}{6} = \frac{180 - 5r}{6}$

Exponent of $5$: Similarly, we combine the exponents of $5$: $(-15 + r/4) + (r/2)$. The common denominator is $4$: $-15 + \frac{r}{4} + \frac{2r}{4} = -15 + \frac{3r}{4} = \frac{3r - 60}{4}$

So, the simplified general term, separating the powers of $x$ and $5$, is: $T_{r+1} = {}^{60}C_r \cdot 5^{\left( \frac{3r - 60}{4} \right)} \cdot x^{\left( \frac{180 - 5r}{6} \right)}$

Tip: Always simplify the individual terms $A$ and $B$ into the form $C \cdot x^p$ before applying the general term formula and raising them to powers. This makes the exponent calculations clearer and reduces the chance of arithmetic errors.

Step 3: Find the value of $r$ for the $x^{10}$ term

Our goal is to find the coefficient of $x^{10}$. Therefore, the exponent of $x$ in the general term must be equal to $10$. We set up an equation to solve for $r$: $\frac{180 - 5r}{6} = 10$ To solve for $r$, first multiply both sides by $6$: $180 - 5r = 60$ Next, subtract $180$ from both sides: $-5r = 60 - 180$ $-5r = -120$ Finally, divide by $-5$: $r = \frac{-120}{-5}$ $r = 24$ We verify that $r=24$ is a valid integer value, as $0 \le r \le n$ (i.e., $0 \le 24 \le 60$).

Step 4: Calculate the coefficient of $x^{10}$

Now that we have the value of $r$, we substitute $r=24$ into the coefficient part of the general term (everything that is not $x$ raised to its power). The coefficient of $x^{10}$ is: ${}^{60}C_{24} \cdot 5^{\left( \frac{3(24) - 60}{4} \right)}$ First, let's calculate the exponent of $5$: $\frac{3 \times 24 - 60}{4} = \frac{72 - 60}{4} = \frac{12}{4} = 3$ So, the coefficient of $x^{10}$ is: ${}^{60}C_{24} \cdot 5^3$ We can express ${}^{60}C_{24}$ using factorials: $\frac{60!}{24! (60-24)!} \cdot 5^3 = \frac{60!}{24! 36!} \cdot 5^3$

Step 5: Determine the value of $k$ using Legendre's Formula

We are given that the coefficient of $x^{10}$ is $5^k \cdot l$, where $l$ is an integer co-prime to $5$. The condition "co-prime to $5$" means that $l$ has no factors of $5$. This is crucial because it implies that $k$ must represent the exact total power of $5$ present in the entire coefficient. To find the exact power of a prime number $p$ that divides $n!$, we use Legendre's Formula: $E_p(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots$ Here, $\lfloor x \rfloor$ denotes the floor function, which gives the greatest integer less than or equal to $x$. This formula helps us find how many times $p$ is a factor in $n!$.

Let's calculate the exponent of $5$ in $60!$, $24!$, and $36!$:

1. Exponent of $5$ in $60!$ ($E_5(60!)$): $E_5(60!) = \left\lfloor \frac{60}{5} \right\rfloor + \left\lfloor \frac{60}{5^2} \right\rfloor + \left\lfloor \frac{60}{5^3} \right\rfloor + \dots$ $E_5(60!) = \left\lfloor 12 \right\rfloor + \left\lfloor \frac{60}{25} \right\rfloor + \left\lfloor \frac{60}{125} \right\rfloor + \dots$ $E_5(60!) = 12 + \left\lfloor 2.4 \right\rfloor + \left\lfloor 0.48 \right\rfloor + \dots$ $E_5(60!) = 12 + 2 + 0 = 14$

2. Exponent of $5$ in $24!$ ($E_5(24!)$): $E_5(24!) = \left\lfloor \frac{24}{5} \right\rfloor + \left\lfloor \frac{24}{5^2} \right\rfloor + \dots$ $E_5(24!) = \left\lfloor 4.8 \right\rfloor + \left\lfloor \frac{24}{25} \right\rfloor + \dots$ $E_5(24!) = 4 + \left\lfloor 0.96 \right\rfloor + \dots$ $E_5(24!) = 4 + 0 = 4$

3. Exponent of $5$ in $36!$ ($E_5(36!)$): $E_5(36!) = \left\lfloor \frac{36}{5} \right\rfloor + \left\lfloor \frac{36}{5^2} \right\rfloor + \dots$ $E_5(36!) = \left\lfloor 7.2 \right\rfloor + \left\lfloor \frac{36}{25} \right\rfloor + \dots$ $E_5(36!) = 7 + \left\lfloor 1.44 \right\rfloor + \dots$ $E_5(36!) = 7 + 1 = 8$

Now, we determine the total exponent of $5$ in the expression $\frac{60!}{24! 36!} \cdot 5^3$. The exponent of $5$ in the combinatorial term ${}^{60}C_{24} = \frac{60!}{24! 36!}$ is found by subtracting the exponents of the denominator from the exponent of the numerator: Exponent of $5$ in ${}^{60}C_{24} = E_5(60!) - E_5(24!) - E_5(36!) = 14 - 4 - 8 = 2$.

So, the coefficient of $x^{10}$ can be expressed as $5^2 \cdot (\text{some integer coprime to 5}) \cdot 5^3$. Combining the powers of $5$: $5^2 \cdot 5^3 = 5^{2+3} = 5^5$ Thus, the coefficient of $x^{10}$ is of the form $5^5 \cdot l'$, where $l'$ is an integer coprime to $5$. Comparing this with the given form $5^k \cdot l$, we conclude that: $k = 5$

Common Mistake: A frequent error when applying Legendre's Formula is incorrectly using the ceiling function ($\lceil x \rceil$) instead of the floor function ($\lfloor x \rfloor$). Always remember to use the floor function for accurate prime factorization.

Summary and Key Takeaway

This problem effectively combines two fundamental concepts in mathematics: the binomial theorem and number theory (specifically, finding the exponent of a prime in a factorial). The solution involves four critical steps: first, correctly setting up and simplifying the general term of the binomial expansion; second, determining the value of 'r' corresponding to the desired power of x; third, calculating the coefficient term involving factorials and constants; and finally, using Legendre's Formula to extract the precise power of a prime (in this case, 5) from the combinatorial part, which is essential due to the condition that $l$ is coprime to $5$. Mastery of these individual concepts and their combined application is crucial for solving such problems.