Final answer:
The number of terms with integer powers in the expansion of (y¹⁄⁵ x¹⁄¹⁰)⁵⁵ is 6, as only terms where the exponent is a multiple of 10 will have integer exponents for both x and y.
Therefore, the correct answer is: A. 56
Step-by-step explanation:
The question asks for the number of terms in the expansion of (y¹⁄⁵ x¹⁄¹⁰)⁵⁵ in which powers of x and y are free from radical signs. To solve this, we observe that when raising a binomial to a power, the number of terms is one more than the power itself.
Given that we have a binomial to the power of 55, the number of terms would usually be 56. However, as we want to ensure that the powers of x and y are integer powers (i.e., free from radical signs), we need to determine which terms meet this criterion.
Each term in the binomial expansion will be in the form of C(n, k) y¹⁄⁵⁵× y¹⁄⁵⁵-k x¹⁄⁵⁵k. We need to find values of k such that both exponents are integers.
Because y is raised to the 1/5 power and x to the 1/10 power, only terms where k is a multiple of 10 will have integer exponents for both x and y after the expansion. The values of k can be 0, 10, 20, ..., up to and including 550/10=55. Hence, there will be 55/10+1=6 terms with integer powers for x and y.
This excludes terms that do not lead to integer powers, and our final count only includes terms with integer exponents.