Math Problem Statement

Find the specified sum for each of the following geometric series and write an expression for the given sequences.

Solution

Let's go through the questions in the image step by step.

1. Find the specified sum for each of the following geometric series:

a. 5+10+20+5 + 10 + 20 + \dots find S10S_{10}

This is a geometric series with:

  • First term a=5a = 5
  • Common ratio r=105=2r = \frac{10}{5} = 2

The sum of the first nn terms of a geometric series is given by the formula: Sn=a(1rn)1r(for r1)S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{(for } r \neq 1) For S10S_{10}, we plug in:

  • a=5a = 5
  • r=2r = 2
  • n=10n = 10 S10=5(1210)12S_{10} = \frac{5(1 - 2^{10})}{1 - 2}

b. 13+91 - 3 + 9 - \dots find S6S_6

This is another geometric series with alternating signs:

  • First term a=1a = 1
  • Common ratio r=3r = -3

Again, using the formula for the sum of the first nn terms: S6=1(1(3)6)1(3)S_6 = \frac{1(1 - (-3)^6)}{1 - (-3)}

c. 43+23+13+\frac{4}{3} + \frac{2}{3} + \frac{1}{3} + \dots find S5S_5

This geometric series has:

  • First term a=43a = \frac{4}{3}
  • Common ratio r=12r = \frac{1}{2}

Applying the sum formula: S5=43(1(12)5)112S_5 = \frac{\frac{4}{3}(1 - \left(\frac{1}{2}\right)^5)}{1 - \frac{1}{2}}

2. For the geometric sequence 2,6,18,2, 6, 18, \dots write an expression for:

a. The nn-th term

In a geometric sequence, the nn-th term is given by: an=arn1a_n = a \cdot r^{n-1} Here, a=2a = 2 and r=62=3r = \frac{6}{2} = 3. So the expression for the nn-th term is: an=23n1a_n = 2 \cdot 3^{n-1}

b. The sum of the first nn terms

Using the sum formula again: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} For this series, a=2a = 2 and r=3r = 3, so: Sn=2(13n)13S_n = \frac{2(1 - 3^n)}{1 - 3}

3. For the geometric sequence 1,12,14,18,1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots, write an expression for:

a. The nn-th term

Here:

  • First term a=1a = 1
  • Common ratio r=12r = \frac{1}{2}

So the nn-th term is: an=1(12)n1a_n = 1 \cdot \left(\frac{1}{2}\right)^{n-1} Simplifying: an=(12)n1a_n = \left(\frac{1}{2}\right)^{n-1}

b. The sum of the first nn terms

Using the sum formula for geometric series: Sn=1(1(12)n)112S_n = \frac{1(1 - \left(\frac{1}{2}\right)^n)}{1 - \frac{1}{2}} Simplifying: Sn=2(1(12)n)S_n = 2 \left(1 - \left(\frac{1}{2}\right)^n\right)


Would you like a detailed breakdown of any particular solution or calculation?

Here are 5 relative questions for further practice:

  1. How would the sum formula change if the common ratio is negative?
  2. Can you find S5S_5 for the sequence 1+2+4+8+16+1 + 2 + 4 + 8 + 16 + \dots?
  3. For the series 2,6,18,54,2, 6, 18, 54, \dots, calculate the sum of the first 7 terms.
  4. How would you calculate the 12th term of the series 3+6+12+24+3 + 6 + 12 + 24 + \dots?
  5. What happens to the sum of an infinite geometric series if r<1|r| < 1?

Tip: When working with geometric series, always check if the common ratio is between -1 and 1, as this allows for the sum of an infinite series to exist.

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Math Problem Analysis

Mathematical Concepts

Geometric Series
Sequences
Summation

Formulas

Sum of the first n terms of a geometric series: S_n = a(1 - r^n) / (1 - r)
n-th term of a geometric sequence: a_n = a * r^(n-1)

Theorems

Geometric series summation

Suitable Grade Level

Grades 9-12