For what value of n, the sum of digits in the number $({10}^n+1)$ is 2?

The correct option is For any whole number n.

Let us analyze the given expression:

$E={10}^n+1$

where $n$ is a whole number (i.e., $n\in \{0,1,2,3,\dots \}$).

Let us evaluate the value of the expression and the sum of its digits for different whole numbers $n$:

Case 1: When $n=0$

${10}^0+1=1+1=2$

The number is $2$, which consists of a single digit $2$.
Sum of digits = $2$.

Case 2: When $n=1$

${10}^1+1=10+1=11$

The digits of $11$ are $1$ and $1$.
Sum of digits = $1+1=2$.

Case 3: When $n=2$

${10}^2+1=100+1=101$

The digits of $101$ are $1$, $0$, and $1$.
Sum of digits = $1+0+1=2$.

General Case: For any non-negative integer (whole number) $n$

For any whole number $n>0$, ${10}^n$ consists of the digit $1$ followed by $n$ zeros (i.e., $100\dots 0$).

Adding $1$ to ${10}^n$ gives a number formed by the digit $1$, followed by $(n-1)$ zeros, and ending with the digit $1$:

${10}^n+1=100\dots 01$

Therefore, the sum of the digits is always:

$\text{Sum of digits}=1+0+0+\dots +0+1=2$

Thus, the sum of the digits in the number $({10}^n+1)$ is equal to $2$ for any whole number $n$.