Math Problem Statement

find the sum of the first 8th term of the geometric sequence 2,4,8,16

Solution

The given geometric sequence is 2,4,8,16,2, 4, 8, 16, \dots, where:

  • First term (aa) = 2
  • Common ratio (rr) = 42=2\frac{4}{2} = 2

The formula to find the sum of the first nn terms of a geometric sequence is:

Sn=a(rn1)r1S_n = a \frac{(r^n - 1)}{r - 1}

For the first 8 terms (n=8n = 8):

S8=2×(281)21S_8 = 2 \times \frac{(2^8 - 1)}{2 - 1}

Let's calculate:

S8=2×(2561)=2×255=510S_8 = 2 \times (256 - 1) = 2 \times 255 = 510

Thus, the sum of the first 8 terms of the sequence is 510510.

Would you like more details on how the formula works, or do you have any other questions?


Here are 5 related questions:

  1. What is the sum of the first 10 terms of this sequence?
  2. How would the sum change if the common ratio was 3?
  3. Can you derive the sum for an infinite geometric series with r=12r = \frac{1}{2}?
  4. What is the product of the first 5 terms of this geometric sequence?
  5. How do you find the sum of an arithmetic sequence?

Tip: In a geometric series, the sum grows rapidly if the common ratio is greater than 1.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences
Series Summation

Formulas

Sum of first n terms of a geometric sequence: Sn = a * (r^n - 1) / (r - 1)

Theorems

Geometric series sum formula

Suitable Grade Level

Grades 9-11