Math Problem Statement

soluciona

Solve the infinite series: 1 + 1/10 + 4/10^2 + 6/10^3 + 1/10^4 + 4/10^5 + 6/10^6 + ...

Solution

The given series appears to follow a repetitive pattern in its digits and denominators. Let’s carefully analyze and calculate its sum.

The series is:

1+110+4102+6103+1104+4105+6106+1107+4108+6109+1 + \frac{1}{10} + \frac{4}{10^2} + \frac{6}{10^3} + \frac{1}{10^4} + \frac{4}{10^5} + \frac{6}{10^6} + \frac{1}{10^7} + \frac{4}{10^8} + \frac{6}{10^9} + \dots

This is an infinite series with repeating digits 1,4,61, 4, 6 in the decimal places. Let’s write it as:

S=1+0.1+0.004+0.0006+0.0001+0.00004+0.000006+S = 1 + 0.1 + 0.004 + 0.0006 + 0.0001 + 0.00004 + 0.000006 + \dots

Step 1: Recognizing the repeating pattern

The digits 1,4,61, 4, 6 repeat every 3 terms. Hence, the series can be grouped as:

S=(1+0.0001+0.0000001+)+(0.1+0.00004+0.000000004+)+(0.004+0.000006+0.000000006+)S = (1 + 0.0001 + 0.0000001 + \dots) + (0.1 + 0.00004 + 0.000000004 + \dots) + (0.004 + 0.000006 + 0.000000006 + \dots)

Each group is a geometric series.

Step 2: Sum of each geometric series

For the first group (1+0.0001+0.0000001+1 + 0.0001 + 0.0000001 + \dots):

  • First term (aa) = 11
  • Common ratio (rr) = 0.00010.0001

The sum of an infinite geometric series is:

Sum=a1r=110.0001=10.99991.0001\text{Sum} = \frac{a}{1 - r} = \frac{1}{1 - 0.0001} = \frac{1}{0.9999} \approx 1.0001

For the second group (0.1+0.00004+0.000000004+0.1 + 0.00004 + 0.000000004 + \dots):

  • First term (aa) = 0.10.1
  • Common ratio (rr) = 0.00040.0004

Sum=a1r=0.110.0004=0.10.99960.10004\text{Sum} = \frac{a}{1 - r} = \frac{0.1}{1 - 0.0004} = \frac{0.1}{0.9996} \approx 0.10004

For the third group (0.004+0.000006+0.000000006+0.004 + 0.000006 + 0.000000006 + \dots):

  • First term (aa) = 0.0040.004
  • Common ratio (rr) = 0.0000010.000001

Sum=a1r=0.00410.0000010.004\text{Sum} = \frac{a}{1 - r} = \frac{0.004}{1 - 0.000001} \approx 0.004

Step 3: Adding the results

Now, add the sums of the three groups:

S1.0001+0.10004+0.0041.10414S \approx 1.0001 + 0.10004 + 0.004 \approx 1.10414

Final Answer:

The sum of the infinite series is approximately:

S=1.10414S = 1.10414

Would you like me to elaborate further or verify any part? Here are some additional questions to explore:

  1. How does the sum change if the repeating pattern changes?
  2. What happens if we truncate the series after a certain number of terms?
  3. How does the calculation change for a different common ratio?
  4. Can this series be expressed as a single decimal expansion?
  5. How would the result vary if the sequence of repeating digits differs?

Tip: When dealing with repeating decimals or series, always check for geometric patterns to simplify calculations.

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Math Problem Analysis

Mathematical Concepts

Infinite series
Geometric progression
Summation of series

Formulas

Sum of an infinite geometric series: S = a / (1 - r)

Theorems

Infinite geometric series formula

Suitable Grade Level

Grades 10-12