1. **State the problem:** We need to evaluate the summation $$\sum_{j=0}^{6} (-1)^j$$ which means adding the values of $(-1)^j$ for $j=0,1,2,\ldots,6$. 2. **Recall the formula and rules:** The term $(-1)^j$ alternates between 1 and -1 depending on whether $j$ is even or odd. 3. **Calculate each term:** - For $j=0$, $(-1)^0 = 1$ - For $j=1$, $(-1)^1 = -1$ - For $j=2$, $(-1)^2 = 1$ - For $j=3$, $(-1)^3 = -1$ - For $j=4$, $(-1)^4 = 1$ - For $j=5$, $(-1)^5 = -1$ - For $j=6$, $(-1)^6 = 1$ 4. **Sum all terms:** $$1 + (-1) + 1 + (-1) + 1 + (-1) + 1 = (1 - 1) + (1 - 1) + (1 - 1) + 1 = 0 + 0 + 0 + 1 = 1$$ 5. **Final answer:** $$\sum_{j=0}^{6} (-1)^j = 1$$