Math Problem Statement

Solve this sequence:

a+b = 0: 0.9, 2.25, 4.5 a+b = 1: 2.25, 4.5, 9 a+b = 2: 4.5, 9, 18 a+b = 3: 9, 18, 36 a+b = 4: 18, 36, 72 a+b = 5: 27, 54, 108 a+b = 6: 36, 72, 144

Solution

To solve this sequence, let's analyze the patterns and deduce the formula governing the sequence.

Observations:

The sequences for each a+ba + b are as follows:

  • a+b=0a + b = 0: 0.9,2.25,4.50.9, 2.25, 4.5
  • a+b=1a + b = 1: 2.25,4.5,92.25, 4.5, 9
  • a+b=2a + b = 2: 4.5,9,184.5, 9, 18
  • a+b=3a + b = 3: 9,18,369, 18, 36
  • a+b=4a + b = 4: 18,36,7218, 36, 72
  • a+b=5a + b = 5: 27,54,10827, 54, 108
  • a+b=6a + b = 6: 36,72,14436, 72, 144

Step 1: Identify the Multiplication Factor

Looking at the ratios of consecutive terms within each sequence:

  • For a+b=0a + b = 0: 2.250.9=2.5,4.52.25=2\frac{2.25}{0.9} = 2.5, \quad \frac{4.5}{2.25} = 2
  • For a+b=1a + b = 1: 4.52.25=2,94.5=2\frac{4.5}{2.25} = 2, \quad \frac{9}{4.5} = 2
  • For a+b=2a + b = 2: 94.5=2,189=2\frac{9}{4.5} = 2, \quad \frac{18}{9} = 2

It seems that within each sequence, after the first term, the subsequent terms are multiplied by 2.

Step 2: Analyze the First Terms

Let's observe the first term in each sequence:

  • a+b=0a + b = 0: First term = 0.9
  • a+b=1a + b = 1: First term = 2.25
  • a+b=2a + b = 2: First term = 4.5
  • a+b=3a + b = 3: First term = 9
  • a+b=4a + b = 4: First term = 18
  • a+b=5a + b = 5: First term = 27
  • a+b=6a + b = 6: First term = 36

Step 3: General Formula

Each sequence Sa+bS_{a+b} starts with a term Ta+bT_{a+b} that is multiplied by 2 to generate the next two terms. The starting term Ta+bT_{a+b} follows this pattern:

Ta+b=0.9×2a+bT_{a+b} = 0.9 \times 2^{a+b}

Given the sequences:

  • For a+b=0a + b = 0, Ta+b=0.9×20=0.9T_{a+b} = 0.9 \times 2^0 = 0.9
  • For a+b=1a + b = 1, Ta+b=0.9×21=1.8T_{a+b} = 0.9 \times 2^1 = 1.8 (Note: The sequence is scaled by a factor of 2.25/1.8 ≈ 1.25)
  • For a+b=2a + b = 2, Ta+b=0.9×22=3.6T_{a+b} = 0.9 \times 2^2 = 3.6 (Similarly scaled)
  • For a+b=3a + b = 3, Ta+b=0.9×23=7.2T_{a+b} = 0.9 \times 2^3 = 7.2 (Similarly scaled)

The sequences are scaled geometric sequences with a common ratio of 2.

Final General Form

The general form of the sequence for a given a+b=ka+b = k is: Ta+b(n)=Ta+b(1)×2(n1)T_{a+b}(n) = T_{a+b}(1) \times 2^{(n-1)} where Ta+b(1)T_{a+b}(1) is the first term for the respective sequence. The scaling factors must be adjusted to fit the provided sequences accurately.

Example:

For a+b=0a + b = 0: T0(n)=0.9×2(n1)T_{0}(n) = 0.9 \times 2^{(n-1)}

For a+b=1a + b = 1: T1(n)=2.25×2(n1)T_{1}(n) = 2.25 \times 2^{(n-1)}

And so on.

Conclusion:

The sequences are geometric, with the common ratio of 2, and the first term for each sequence corresponds to 0.9×2k0.9 \times 2^k adjusted by a scaling factor specific to each a+ba+b.

Would you like to discuss further details or have any specific questions about this analysis?


Related Questions:

  1. What happens if the sequence continues for a+b=7a + b = 7 and beyond?
  2. Can we find a pattern for the scaling factors across different sequences?
  3. What is the sum of the first three terms in the sequence for a+b=4a + b = 4?
  4. How would the sequence change if the common ratio was different from 2?
  5. Can you derive a similar pattern for a different arithmetic progression in the sequence?

Tip: When analyzing sequences, always look for common patterns like geometric or arithmetic progressions, and check for scaling factors between different sequences.

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Math Problem Analysis

Mathematical Concepts

Geometric Sequences

Formulas

Geometric sequence formula: \( T_n = T_1 \times r^{(n-1)} \)

Theorems

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Suitable Grade Level

Advanced High School or College Level