sum _(k=0)^1(2-k)= square 7 8 4 9 5 6 3

This is a simple summation problem. The sigma notationsum$, is a way of expressing a long sum in a concise way. Here, it is summing the expression (2-k) for k from 0 to 1. <br /><br />To solve this, we substitute the values of k into the expression and add them up:<br /><br />When k = 0, the expression (2-k) becomes 2 - 0 = 2.<br />When k = 1, the expression (2-k) becomes 2 - 1 = 1.<br /><br />So, the sum $\sum _{k=0}^{1}(2-k)$ is equal to 2 + 1 = 3.