To find the coefficients of xn in the expansion of (1 + 2x + 3x^2 + ...)^1/2, we can use the binomial theorem. The binomial theorem states that for any real number a and b and any positive integer n, the expansion of (a + b)^n can be written as:

(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n-1) * a^1 * b^(n-1) + C(n, n) * a^0 * b^n

where C(n, r) is the binomial coefficient, also called "n choose r", and is given by the formula:

C(n, r) = n! / (r! * (n-r)!)

In the given expression (1 + 2x + 3x^2 + ...)^1/2, we have a = 1 and b = 2x + 3x^2 + ... Therefore, applying the binomial theorem, the expansion becomes:

(1 + 2x + 3x^2 + ...)^1/2 = C(1/2, 0) * 1^(1/2) * (2x + 3x^2 + ...)^0 + C(1/2, 1) * 1^(1/2 - 1) * (2x + 3x^2 + ...)^1/2 + ...

Simplifying this expression, we find that the coefficient of xn is given by:

C(1/2, n) * 1^(1/2 - n) * (2x + 3x^2 + ...)^n/2

Since C(1/2, n) = 1 for all values of n, and 1^(1/2 - n) = 1, the coefficient of xn simplifies to:

(2x + 3x^2 + ...)^n/2

From this, we can see that the coefficient of xn is simply the expression (2x + 3x^2 + ...)^n/2. Since this expression does not depend on n, the coefficient of xn is constant and equal to 1.

Therefore, the correct answer is option A) 1.