Let's go over the problems in this image, which are focused on evaluating infinite geometric series.

Formula for the sum of an infinite geometric series:

For an infinite geometric series with the first term $a_1$ and common ratio $r$, the sum $S$ is given by: $S = \frac{a_1}{1 - r}, \quad \text{if} \ |r| < 1$ If $|r| \geq 1$, the series does not converge, so no sum can be found.

Solutions to the problems:

1. $81 - 27 + 9 - 3 + \dots$:

   • First term $a_1 = 81$
   • Common ratio $r = \frac{-27}{81} = -\frac{1}{3}$ $S = \frac{81}{1 - (-\frac{1}{3})} = \frac{81}{1 + \frac{1}{3}} = \frac{81}{\frac{4}{3}} = 81 \times \frac{3}{4} = 60.75$

2. $1 + 0.5 + 0.25 + 0.125 + \dots$:

   • First term $a_1 = 1$
   • Common ratio $r = \frac{0.5}{1} = 0.5$ $S = \frac{1}{1 - 0.5} = \frac{1}{0.5} = 2$

3. $3 + 6 + 12 + 24 + \dots$:

   • First term $a_1 = 3$
   • Common ratio $r = \frac{6}{3} = 2$
   • Since $r = 2$, the series diverges because $|r| \geq 1$. No sum exists.

4. $a_1 = -1, r = 3$:

   • The series diverges because $|r| \geq 1$. No sum exists.

5. $a_1 = -3, r = 4$:

   • The series diverges because $|r| \geq 1$. No sum exists.

6. $a_1 = 4, r = -\frac{3}{4}$:

   • First term $a_1 = 4$
   • Common ratio $r = -\frac{3}{4}$ $S = \frac{4}{1 - (-\frac{3}{4})} = \frac{4}{1 + \frac{3}{4}} = \frac{4}{\frac{7}{4}} = 4 \times \frac{4}{7} = \frac{16}{7}$

7. $a_1 = 5.5, r = 0.5$:

   • First term $a_1 = 5.5$
   • Common ratio $r = 0.5$ $S = \frac{5.5}{1 - 0.5} = \frac{5.5}{0.5} = 11$

8. Summation $\sum_{i=1}^{\infty} \left( \frac{1}{3} \right)^{i-1}$:

   • This is a geometric series with $a_1 = 1$ (first term is $\left(\frac{1}{3}\right)^0 = 1$) and $r = \frac{1}{3}$ $S = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

9. Summation $\sum_{i=1}^{\infty} 2^{i-1}$:

   • This is a geometric series with $a_1 = 1$ (first term is $2^0 = 1$) and $r = 2$
   • Since $r = 2$, the series diverges because $|r| \geq 1$. No sum exists.

10. Summation $\sum_{i=1}^{\infty} 4^{i-1}$:

    • This is a geometric series with $a_1 = 1$ (first term is $4^0 = 1$) and $r = 4$
    • Since $r = 4$, the series diverges because $|r| \geq 1$. No sum exists.

Would you like further details or clarifications?

5 Related Questions:

1. What conditions are needed for a geometric series to converge?
2. How does the common ratio affect the sum of an infinite series?
3. Can geometric series with a common ratio greater than 1 ever have a finite sum?
4. How can you express a recurring decimal as an infinite geometric series?
5. What is the significance of geometric series in real-life applications?

Tip:

When the common ratio $|r|$ is less than 1, the series converges to a finite value, but when $|r| \geq 1$, the series diverges (no finite sum).