Math Problem Statement

Find the sum of the first 11 terms of the geometric sequence 4, 8, 16, 32,...

Solution

This problem is asking to find the sum of the first 11 terms of a geometric sequence, where the sequence begins with:

4,8,16,32,4, 8, 16, 32, \dots

Step 1: Identify the values in the geometric sequence

  • The first term a=4a = 4
  • The common ratio r=84=2r = \frac{8}{4} = 2

Step 2: Use the formula for the sum of the first nn terms of a geometric sequence

The sum SnS_n of the first nn terms of a geometric sequence is given by:

Sn=a1rn1rforr1S_n = a \frac{1 - r^n}{1 - r} \quad \text{for} \quad r \neq 1

Where:

  • a=4a = 4 is the first term,
  • r=2r = 2 is the common ratio,
  • n=11n = 11 is the number of terms.

Step 3: Substitute the known values into the formula

S11=4×121112S_{11} = 4 \times \frac{1 - 2^{11}}{1 - 2}

Let's calculate this step-by-step. I'll perform the calculation next.The sum of the first 11 terms of the given geometric sequence is 81888188.

Would you like further details or have any additional questions?

Here are 5 related questions to explore:

  1. How is the formula for the sum of a geometric sequence derived?
  2. What happens to the sum when the common ratio is less than 1?
  3. How would the sum change if we were asked for only the first 5 terms?
  4. Can the sum of an infinite geometric series be calculated if r>1r > 1?
  5. What are some real-life applications of geometric sequences?

Tip: When calculating geometric series, always check if the ratio rr is greater than 1 or between 0 and 1, as it affects the series' behavior.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequence
Summation

Formulas

S_n = a * (1 - r^n) / (1 - r) for r ≠ 1

Theorems

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Suitable Grade Level

Grades 9-12