Sum of digits of 9-digit repunit squared

A 9-digit number consisting entirely of ones is a repunit, written as 111,111,111. Squaring an n-digit repunit (where n \le 9) generates a perfect palindromic structure string running from 1 up to n and back down to 1.

For n = 9, the squared result is: 12345678987654321.

Calculate the sum of the digits of this resulting number: Sum = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) × 2 + 9 Sum = 36 × 2 + 9 = 72 + 9 = 81. Notice a shortcut identity: The sum of digits for the square of an ⟨MATH⟩n⟨/MATH⟩-digit repunit is always exactly ⟨MATH⟩n²⟨/MATH⟩. For n = 9, Sum = 9² = 81.

Answer: (c).