e) The series can be represented in summation notation as:
\[\sum_{n=1}^{\infty} \frac{162}{3^{n-1}}\]
f) To find the sum of this infinite series, we can use the formula for the sum of an infinite geometric series:
\[S = \frac{a}{1 - r}\]
Where:
a = first term = 162
r = common ratio = 1/3
So, plugging in the values:
\[S = \frac{162}{1 - 1/3}\]
\[S = \frac{162}{2/3}\]
\[S = 162 \times \frac{3}{2}\]
\[S = 243\]
Therefore, the sum of the infinite series 162 + 54 + 18 + 6 + … is 243.
For the infinite series 162 + 54 + 18 + 6 + …
e) Write the series in summation notation where the lower limit is n=1.
f) What is the sum of this infinite series? Reminder: You will be graded for showing the correct process as well as the correct answer using algebraic work.
1 answer