Question

If the coefficient of x10 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∈N and l is co-prime to 5, then k is equal to ___________.

Answer 1

Correct Answer: 5

To find the coefficient of x¹⁰ in the binomial expansion \(\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}} + \frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}\), we use the binomial theorem:
For \((a+b)^n, the term containing x is given by \(\binom{n}{r} a^{n-r} b^r\).

Set \(a=\frac{\sqrt{x}}{5^{1/4}}\) and \(b=\frac{\sqrt{5}}{x^{1/3}}\), then expand:

The term will be:
\(\binom{60}{r} \left(\frac{\sqrt{x}}{5^{1/4}}\right)^{60-r} \left(\frac{\sqrt{5}}{x^{1/3}}\right)^r\)
which simplifies to:
\(\binom{60}{r} \frac{x^{(60-r)/2}}{5^{(60-r)/4}} \cdot \frac{5^{r/2}}{x^{r/3}}\).

Combine and simplify:
\(\rightarrow \binom{60}{r} \cdot x^{(60-r)/2 - r/3} \cdot 5^{r/2 - (60-r)/4}\)

Equating the powers for x¹⁰:
\(\frac{60-r}{2} - \frac{r}{3} = 10\)

Multiply the entire equation by 6 to clear fractions:
\(3(60-r) - 2r = 60\)
Simplifies to:
\(180 - 3r - 2r = 60\)
\(180 - 5r = 60 \rightarrow 5r = 120 \rightarrow r = 24\)

Now, substitute r=24 for coefficient:
\(5^{24/2 - (60-24)/4} = 5^{12-9} = 5^3\)

Thus, k = 3. Co-prime factor l exists for k=3 according to condition.
Final: k = 3, which fits the given range (5, 5).