The sum of digits in the number (10ⁿ + 1) is 2 for any whole number n.

If n = 0, then 10⁰ + 1 = 1 + 1 = 2. The sum of digits is 2.
If n is a positive integer (n > 0), the number 10ⁿ consists of the digit '1' followed by 'n' zeros (e.g., 10¹ = 10, 10² = 100, 10³ = 1000).
Therefore, (10ⁿ + 1) will consist of the digit '1' at the (n+1)^th place (from the right, starting at 0) and the digit '1' at the units place, with 'n-1' zeros in between (e.g., 10¹ + 1 = 11; 10² + 1 = 101; 10³ + 1 = 1001).
The sum of the digits in such a number will always be 1 + 0 + ... + 0 + 1 = 2.
This condition holds true for all whole numbers (0, 1, 2, 3, ...).

Option A (For n = 0 only): This is incorrect because the condition also holds for all positive integers (e.g., n=1, 2, 3, ...), not just for n=0.
Option C (For any positive integer n only): This is incorrect because it excludes n=0, for which the sum of digits in (10⁰ + 1) is 2.
Option D (For any real number n): This is incorrect. The concept of "sum of digits" is typically applied to integers or numbers with a finite decimal representation. For non-integer real values of n (e.g., n=0.5), 10ⁿ + 1 would result in a non-integer value (e.g., 10^0.5 + 1 = √10 + 1 ≈ 4.1622...), for which the sum of digits is not 2 and the definition of 'sum of digits' becomes ambiguous in the context of the question.