Math Problem Statement

a₁, a₂, a₃, ... aₙ is a geometric sequence with a₂ = 4/25, a₄ = 16/625, and r > 0. Find the sum of the infinite geometric series ∑ₙ₌₁ aₙ.

Solution

We are tasked with finding the sum of an infinite geometric sequence. Let us break it down step by step.

Problem Breakdown:

  1. Given:

    • a2=425a_2 = \frac{4}{25}
    • a4=16625a_4 = \frac{16}{625}
    • r>0r > 0
  2. Geometric sequence relationship:

    • Each term is related by the common ratio rr: an+1=rana_{n+1} = r \cdot a_n
  3. Find the sum of the infinite geometric series:

    • The formula for the sum of an infinite geometric sequence is: S=a11r,where r<1.S = \frac{a_1}{1 - r}, \quad \text{where } |r| < 1.

Step 1: Find the common ratio rr

From the given terms a2a_2 and a4a_4: a4=a2r2a_4 = a_2 \cdot r^2 Substituting a2=425a_2 = \frac{4}{25} and a4=16625a_4 = \frac{16}{625}: 16625=425r2\frac{16}{625} = \frac{4}{25} \cdot r^2 Solve for r2r^2: r2=16625425=16625254=16256254=4002500=15.r^2 = \frac{\frac{16}{625}}{\frac{4}{25}} = \frac{16}{625} \cdot \frac{25}{4} = \frac{16 \cdot 25}{625 \cdot 4} = \frac{400}{2500} = \frac{1}{5}. r=15=15.r = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}.


Step 2: Find a1a_1

We know a2=ra1a_2 = r \cdot a_1. Substituting r=15r = \frac{1}{\sqrt{5}} and a2=425a_2 = \frac{4}{25}: 425=15a1\frac{4}{25} = \frac{1}{\sqrt{5}} \cdot a_1 a1=42515=4255=4525.a_1 = \frac{\frac{4}{25}}{\frac{1}{\sqrt{5}}} = \frac{4}{25} \cdot \sqrt{5} = \frac{4\sqrt{5}}{25}.


Step 3: Find the sum of the infinite geometric sequence

Using the sum formula: S=a11r,S = \frac{a_1}{1 - r}, substitute a1=4525a_1 = \frac{4\sqrt{5}}{25} and r=15r = \frac{1}{\sqrt{5}}: S=4525115.S = \frac{\frac{4\sqrt{5}}{25}}{1 - \frac{1}{\sqrt{5}}}.

Simplify the denominator 1151 - \frac{1}{\sqrt{5}}: 115=515.1 - \frac{1}{\sqrt{5}} = \frac{\sqrt{5} - 1}{\sqrt{5}}.

The sum becomes: S=4525515=4525551=4525(51)=2025(51)=45(51).S = \frac{\frac{4\sqrt{5}}{25}}{\frac{\sqrt{5} - 1}{\sqrt{5}}} = \frac{4\sqrt{5}}{25} \cdot \frac{\sqrt{5}}{\sqrt{5} - 1} = \frac{4 \cdot 5}{25(\sqrt{5} - 1)} = \frac{20}{25(\sqrt{5} - 1)} = \frac{4}{5(\sqrt{5} - 1)}.

Rationalize the denominator: 45(51)5+15+1=4(5+1)5((5)212)=4(5+1)5(51)=4(5+1)20.\frac{4}{5(\sqrt{5} - 1)} \cdot \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{4(\sqrt{5} + 1)}{5((\sqrt{5})^2 - 1^2)} = \frac{4(\sqrt{5} + 1)}{5(5 - 1)} = \frac{4(\sqrt{5} + 1)}{20}. Simplify further: S=5+15.S = \frac{\sqrt{5} + 1}{5}.


Final Answer:

S=5+15\boxed{S = \frac{\sqrt{5} + 1}{5}}

Would you like further clarification or details? Here are some related questions:

  1. What are the general properties of geometric sequences?
  2. How is the formula for the sum of an infinite series derived?
  3. How does the value of rr affect the convergence of the series?
  4. Can this method be extended to find partial sums of the sequence?
  5. What is the impact of rationalizing the denominator in mathematical computations?

Tip: Always ensure r<1|r| < 1 before attempting to compute the sum of an infinite geometric series.

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Math Problem Analysis

Mathematical Concepts

Sequences and Series
Geometric Sequences
Infinite Series
Common Ratio

Formulas

aₙ₊₁ = r * aₙ (geometric sequence relation)
Sum of infinite geometric series: S = a₁ / (1 - r)

Theorems

Convergence of geometric series (if |r| < 1)

Suitable Grade Level

Grades 10-12