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OABC is a square where O is the origin and AB = 1. Consider the set of points $$s = {(x_{i},y_{i})}$$ in the square such that $$x_{i}+y_{i}$$≤1. Let $$P (x_{1}, y_{1})$$ and $$Q (x_{2}, y_{2})$$ be two such points. Two operations addition (+) and multiplication (.) on S are defined as
$$P + Q = (x_{1}+x_{2} - x_{1}x_{2},y_{1}y_{2})$$
$$P.Q = (x_{1}x_{2},y_{1}+y_{2} - y_{1}y_{2})$$
Let $$P=\left(x_1,\ y_1\right)$$ and $$Q=\left(x_2,\ y_2\right)$$
$$P^2=P\times P=(x_1\times x_1, y_1+y_1-y_1\times y_1)=(x_1^2, y_1(2-y_1))$$
Since $$x_1$$ and $$y_1$$ are both less than 1, $$x_1^2<x_1$$ and $$y_1(2-y_1)>y_1$$
As P is raised to higher powers, the x-coordinate will keep on decreasing and the y-coordinate will keep on increasing.
As the value of n tends to infinity, $$x_1$$ and $$y_1$$ will tend towards 0 and 1, resptectively.
Thus, $$P^n=(0, 1)$$.
Similarly, $$Q^n=(0, 1)$$
Thus, $$P^n+Q^n=(0\times 0, 1+1-1\times 1)=(0, 1)$$
Hence, the answer is option C.
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