Question: User: 3n = 2,73
e) Sā of 3/5 + 9/10 + 27/20 + 81/40 + ...
3/5 (1 - (3/2)^6) / (1 - 3/2)
1. **State the problem:** We want to find the sum of the first 6 terms, $S_6$, of the series $\frac{3}{5} + \frac{9}{10} + \frac{27}{20} + \frac{81}{40} + \dots$.
2. **Identify the type of series:** This is a geometric series where each term is multiplied by a common ratio $r$ to get the next term.
3. **Find the first term $a$ and common ratio $r$:**
- The first term $a = \frac{3}{5}$.
- To find $r$, divide the second term by the first term:
$$r = \frac{\frac{9}{10}}{\frac{3}{5}} = \frac{9}{10} \times \frac{5}{3} = \frac{45}{30} = \frac{3}{2}.$$
4. **Formula for the sum of the first $n$ terms of a geometric series:**
$$S_n = a \times \frac{1 - r^n}{1 - r}$$
5. **Apply the formula for $n=6$:**
$$S_6 = \frac{3}{5} \times \frac{1 - \left(\frac{3}{2}\right)^6}{1 - \frac{3}{2}}$$
6. **Calculate $\left(\frac{3}{2}\right)^6$:**
$$\left(\frac{3}{2}\right)^6 = \frac{3^6}{2^6} = \frac{729}{64}$$
7. **Substitute back:**
$$S_6 = \frac{3}{5} \times \frac{1 - \frac{729}{64}}{1 - \frac{3}{2}} = \frac{3}{5} \times \frac{\frac{64}{64} - \frac{729}{64}}{\frac{2}{2} - \frac{3}{2}} = \frac{3}{5} \times \frac{\frac{64 - 729}{64}}{\frac{2 - 3}{2}} = \frac{3}{5} \times \frac{\frac{-665}{64}}{\frac{-1}{2}}$$
8. **Simplify the complex fraction:**
$$S_6 = \frac{3}{5} \times \frac{-665}{64} \times \frac{2}{-1} = \frac{3}{5} \times \frac{-665 \times 2}{64 \times -1} = \frac{3}{5} \times \frac{1330}{64}$$
9. **Cancel common factors:**
$$\frac{3}{5} \times \frac{1330}{64} = \frac{3 \times 1330}{5 \times 64} = \frac{3990}{320}$$
10. **Simplify the fraction:**
Divide numerator and denominator by 10:
$$\frac{3990}{320} = \frac{399}{32}$$
11. **Final answer:**
$$S_6 = \frac{399}{32} \approx 12.46875$$