If the coefficient of x¹⁰ in the binomial expansion of ( √x/5^(1/4) + √5/x^(1/3) )⁶⁰ is 5ᵏ . l, where l, k ∈ N and l is co-prime to 5, then k is equal to _____________.

Solution

Maths sir · 4 steps

Find r from the power of x first, then count the 5s in both the terms and the binomial coefficient.

1. Tᵣ₊₁ = C(60, r)(x^(1/2)/5^(1/4))⁶⁰⁻ʳ(5^(1/2)/x^(1/3))ʳ

2. Find r. The power of x is (60-r)/2 - r/3. Set it equal to 10: 3(60-r) - 2r = 60, so 180 - 5r = 60 and r = 24.
3. Fives from the terms. The power of 5 from the two terms is -(60-r)/4 + r/2 = -36/4 + 24/2 = -9 + 12 = 3. So the coefficient is C(60, 24) × 5³.
4. Fives in the coefficient. Count the 5s in each factorial: 60! has 12 + 2 = 14, 24! has 4, and 36! has 7 + 1 = 8. So C(60, 24) = 60!/(24! 36!) contains 5¹⁴⁻⁴⁻⁸ = 5², and what is left over is co-prime to 5.
5. So. The coefficient is 5³ × 5² × l = 5⁵ × l with l co-prime to 5, so k = 5.

The common slip: 3. Counting only the 5³ from the terms and treating C(60, 24) as co-prime to 5 gives 3. It still hides 5².

The idea to keep. For the exact power of a prime in a coefficient, count it inside C(n, r) too, using ⌊ n/p ⌋ + ⌊ n/p² ⌋ + … for each factorial.

Answer from the NTA final answer key. Solution by PrepWiser.