If 11+103+1005+... n terms = (10^(n+1) + x n^2 + y) / 9, then the value of x-y is A -19 B 19 C 1 D -1

If $11+103+1005+\cdots n$ terms $= \frac{10^{n+1}+x n^{2}+y}{9}$, then the value of $x-y$ is

A -19 B 19 C 1 D -1

The correct option is B 19.

Let's denote the sum of the series to $n$ terms as $S_n$: $$S_n = 11 + 103 + 1005 + \cdots \text{ (up to $n$ terms)}$$.

Now, let's break down the series: $$ S_n = (10 + 1) + (100 + 3) + (1000 + 5) + \cdots + \left(10^n + (2n - 1)\right) $$

This can be separated into two distinct sums: $$ S_n = \left(10 + 10^2 + 10^3 + \cdots + 10^n\right) + \left(1 + 3 + 5 + \cdots + (2n - 1)\right) $$

1. Sum of the geometric series $10 + 10^2 + 10^3 + \cdots + 10^n$: $$10 \times \frac{(10^n - 1)}{10 - 1} = \frac{10}{9}(10^n - 1)$$

2. Sum of the arithmetic series $1 + 3 + 5 + \cdots + (2n - 1)$: The given sequence is the sum of the first $n$ odd numbers, which equals $n^2$.

Combining these two, we get: $$ S_n = \frac{10}{9}(10^n - 1) + n^2 $$

Next, we can simplify it further: $$ S_n = \frac{10^{n+1} - 10}{9} + n^2 $$ By making common denominators, the expression becomes: $$ S_n = \frac{10^{n+1} - 10 + 9n^2}{9} $$

Thus, comparing the given information, we identify: $$ \frac{10^{n+1} + x n^2 + y}{9} = \frac{10^{n+1} - 10 + 9n^2}{9} $$

By comparing coefficients, we find: $$ x = 9 \quad \text{and} \quad y = -10 $$

Therefore, $$ x - y = 9 - (-10) = 19 $$

So, the correct option is B 19.