1. The problem involves evaluating four different summations with alternating signs and powers of 2. 2. The general form of the summations is $$\sum (-1)^n 2^{f(n)}$$ where $f(n)$ varies. 3. Let's evaluate each summation step-by-step. **A.** $$\sum_{n=1}^6 (-1)^n 2^{-2}$$ - Since $2^{-2} = \frac{1}{4}$ is constant, the sum is $$\frac{1}{4} \sum_{n=1}^6 (-1)^n$$ - The terms of $(-1)^n$ for $n=1$ to $6$ are: $-1, 1, -1, 1, -1, 1$ - Summing these: $(-1 + 1) + (-1 + 1) + (-1 + 1) = 0$ - Therefore, sum A = $$\frac{1}{4} \times 0 = 0$$ **B.** $$\sum_{n=0}^6 (-1)^n 2^{-n}$$ - This is a finite geometric series with first term $a = (-1)^0 2^0 = 1$ - Common ratio $r = -\frac{1}{2}$ - Number of terms $N=7$ - Sum formula: $$S = a \frac{1-r^N}{1-r} = 1 \times \frac{1 - (-\frac{1}{2})^7}{1 + \frac{1}{2}} = \frac{1 - (-\frac{1}{128})}{\frac{3}{2}} = \frac{1 + \frac{1}{128}}{\frac{3}{2}} = \frac{\frac{129}{128}}{\frac{3}{2}} = \frac{129}{128} \times \frac{2}{3} = \frac{258}{384} = \frac{43}{64}$$ **C.** $$\sum_{n=1}^6 (-1)^n 2^n$$ - Calculate each term: - $n=1$: $-1 \times 2 = -2$ - $n=2$: $1 \times 4 = 4$ - $n=3$: $-1 \times 8 = -8$ - $n=4$: $1 \times 16 = 16$ - $n=5$: $-1 \times 32 = -32$ - $n=6$: $1 \times 64 = 64$ - Sum: $-2 + 4 - 8 + 16 - 32 + 64 = 42$ **D.** $$\sum_{n=0}^6 (-1)^n 2^n$$ - This is a finite geometric series with $a=1$, $r=-2$, $N=7$ - Sum formula: $$S = a \frac{1-r^N}{1-r} = 1 \times \frac{1 - (-2)^7}{1 - (-2)} = \frac{1 - (-128)}{1 + 2} = \frac{1 + 128}{3} = \frac{129}{3} = 43$$ **Final answers:** - A = 0 - B = $\frac{43}{64}$ - C = 42 - D = 43