Math Problem Statement

What is the 6th term of the series: ∑_{n=1}^∞ 15(1/2)^n?

Solution

To solve the first question, we are tasked with finding the 6th term of the series:

n=115(12)n\sum_{n=1}^{\infty} 15 \left(\frac{1}{2}\right)^n

Solution

  1. General Term of the Series: The general term for the series is: an=15(12)na_n = 15 \left(\frac{1}{2}\right)^n

  2. Find the 6th Term (n=6n = 6): Substitute n=6n = 6 into the formula: a6=15(12)6a_6 = 15 \left(\frac{1}{2}\right)^6

  3. Simplify: (12)6=164\left(\frac{1}{2}\right)^6 = \frac{1}{64} So: a6=15164=1564a_6 = 15 \cdot \frac{1}{64} = \frac{15}{64}

Answer:

The 6th term of the series is:

1564\boxed{\frac{15}{64}}

Would you like assistance with the second question or any further clarification?


Related Questions:

  1. How do you determine the convergence of geometric series?
  2. Can you derive a general formula for the sum of a finite geometric series?
  3. What happens to the series if the ratio is greater than 1?
  4. How does the convergence test work for infinite series like 1np\frac{1}{n^p}?
  5. Could you compute other terms in the series using the same approach?

Tip: In geometric series, always verify if the ratio lies between 1-1 and 11 to determine convergence.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometric Series
Exponential Functions

Formulas

General term of a geometric series: a_n = ar^n

Theorems

Geometric series term formula

Suitable Grade Level

Grades 11-12 (AP Calculus BC or equivalent)