1. **State the problem:** Find the sum of the arithmetic series: $$2 - \frac{4}{3} + \frac{8}{9} - \frac{16}{27} + \ldots - \frac{4096}{177147}$$ 2. **Identify the type of series:** The given series is not arithmetic because the difference between terms is not constant. Instead, it is a geometric series where each term is multiplied by a common ratio to get the next term. 3. **Find the first term ($a$) and common ratio ($r$):** - First term $a = 2$ - To find $r$, divide the second term by the first term: $$r = \frac{-\frac{4}{3}}{2} = -\frac{4}{3} \times \frac{1}{2} = -\frac{2}{3}$$ 4. **Confirm the common ratio:** - Check the third term: $$2 \times \left(-\frac{2}{3}\right)^2 = 2 \times \frac{4}{9} = \frac{8}{9}$$ - This matches the third term, so $r = -\frac{2}{3}$ is correct. 5. **Find the number of terms ($n$):** - The last term is $-\frac{4096}{177147}$. - The $n$th term of a geometric series is: $$a_n = a \times r^{n-1}$$ - Substitute known values: $$-\frac{4096}{177147} = 2 \times \left(-\frac{2}{3}\right)^{n-1}$$ - Divide both sides by 2: $$\frac{-\frac{4096}{177147}}{2} = \left(-\frac{2}{3}\right)^{n-1}$$ $$-\frac{4096}{177147} \times \frac{1}{2} = -\frac{2048}{177147} = \left(-\frac{2}{3}\right)^{n-1}$$ 6. **Express numerator and denominator as powers:** - $2048 = 2^{11}$ - $177147 = 3^{11}$ So: $$\left(-\frac{2}{3}\right)^{n-1} = -\frac{2^{11}}{3^{11}} = -\left(\frac{2}{3}\right)^{11}$$ 7. **Equate powers:** $$\left(-\frac{2}{3}\right)^{n-1} = -\left(\frac{2}{3}\right)^{11}$$ Since the base is negative, the negative sign outside means the exponent $n-1$ must be odd to produce a negative result. Therefore: $$n - 1 = 11 \implies n = 12$$ 8. **Use the sum formula for geometric series:** $$S_n = a \times \frac{1 - r^n}{1 - r}$$ Substitute $a=2$, $r = -\frac{2}{3}$, and $n=12$: $$S_{12} = 2 \times \frac{1 - \left(-\frac{2}{3}\right)^{12}}{1 - \left(-\frac{2}{3}\right)}$$ 9. **Calculate $r^{12}$:** Since $12$ is even, $\left(-\frac{2}{3}\right)^{12} = \left(\frac{2}{3}\right)^{12}$ $$\left(\frac{2}{3}\right)^{12} = \frac{2^{12}}{3^{12}} = \frac{4096}{531441}$$ 10. **Calculate denominator:** $$1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$$ 11. **Calculate numerator:** $$1 - \frac{4096}{531441} = \frac{531441 - 4096}{531441} = \frac{527345}{531441}$$ 12. **Calculate sum:** $$S_{12} = 2 \times \frac{\frac{527345}{531441}}{\frac{5}{3}} = 2 \times \frac{527345}{531441} \times \frac{3}{5}$$ 13. **Simplify:** $$S_{12} = 2 \times \frac{527345 \times 3}{531441 \times 5} = 2 \times \frac{1582035}{2657205}$$ $$S_{12} = \frac{3164070}{2657205}$$ 14. **Simplify fraction by dividing numerator and denominator by their GCD (105):** $$\cancel{\frac{3164070}{105}} = 30134, \quad \cancel{\frac{2657205}{105}} = 25307$$ So: $$S_{12} = \frac{30134}{25307}$$ 15. **Final answer:** The sum of the series is: $$\boxed{\frac{30134}{25307}}$$