1. The problem is to express the series $1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} + \frac{1}{64}$ as a summation. 2. This is a geometric series where the first term $a = 1$ and the common ratio $r = -\frac{1}{2}$. 3. The general form of a geometric series is: $$ S_n = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^k $$ 4. Here, $a = 1$, $r = -\frac{1}{2}$, and the number of terms $n = 7$ (since there are 7 terms). 5. Therefore, the summation form is: $$ \sum_{k=0}^{6} \left(-\frac{1}{2}\right)^k $$ 6. This summation represents the original series exactly. 7. To find the sum, use the formula for the sum of the first $n$ terms of a geometric series: $$ S_n = a \frac{1-r^n}{1-r} $$ 8. Substitute the values: $$ S_7 = 1 \times \frac{1 - \left(-\frac{1}{2}\right)^7}{1 - \left(-\frac{1}{2}\right)} = \frac{1 - \left(-\frac{1}{2}\right)^7}{1 + \frac{1}{2}} = \frac{1 - \left(-\frac{1}{2}\right)^7}{\frac{3}{2}} $$ 9. Calculate $\left(-\frac{1}{2}\right)^7 = -\frac{1}{128}$. 10. Substitute back: $$ S_7 = \frac{1 - \left(-\frac{1}{128}\right)}{\frac{3}{2}} = \frac{1 + \frac{1}{128}}{\frac{3}{2}} = \frac{\frac{128}{128} + \frac{1}{128}}{\frac{3}{2}} = \frac{\frac{129}{128}}{\frac{3}{2}} $$ 11. Simplify the division: $$ S_7 = \frac{129}{128} \times \frac{2}{3} = \frac{258}{384} = \frac{43}{64} $$ 12. Final answer: $$ \sum_{k=0}^{6} \left(-\frac{1}{2}\right)^k = \frac{43}{64} $$