1. **State the problem:** We need to find the approximate sum of the first 50 terms of two geometric sequences. 2. **Recall the formula for the sum of the first $n$ terms of a geometric sequence:** $$s = a \frac{1-r^n}{1-r}$$ where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. 3. **Sequence a:** $1, \frac{1}{2}, \frac{1}{4}, \ldots$ - Here, $a=1$, $r=\frac{1}{2}$, and $n=50$. - Substitute into the formula: $$s = 1 \times \frac{1-(\frac{1}{2})^{50}}{1-\frac{1}{2}}$$ - Simplify the denominator: $$1-\frac{1}{2} = \frac{1}{2}$$ - So, $$s = \frac{1-(\frac{1}{2})^{50}}{\frac{1}{2}}$$ - Use the cancellation notation: $$s = \frac{1-\cancel{(\frac{1}{2})^{50}}}{\cancel{\frac{1}{2}}}$$ - Actually, we cannot cancel powers here, so we keep it as is. - Calculate $(\frac{1}{2})^{50}$ which is very close to 0. - Therefore, $$s \approx \frac{1-0}{\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$$ 4. **Sequence b:** $1, 3, 9, \ldots$ - Here, $a=1$, $r=3$, and $n=50$. - Substitute into the formula: $$s = 1 \times \frac{1-3^{50}}{1-3}$$ - Simplify the denominator: $$1-3 = -2$$ - So, $$s = \frac{1-3^{50}}{-2}$$ - Use cancellation notation: $$s = \frac{\cancel{1}-3^{50}}{\cancel{-2}}$$ - Since $3^{50}$ is a very large number, $1$ is negligible. - So, $$s \approx \frac{-3^{50}}{-2} = \frac{3^{50}}{2}$$ 5. **Final answers:** - For sequence a, the sum of the first 50 terms is approximately $2$. - For sequence b, the sum of the first 50 terms is approximately $\frac{3^{50}}{2}$, a very large number.