Math Problem Statement

If u_{n+1} = 1/5 u_n, where u_1 = 5, find the exact value of ∑_{r=1}^{∞} u_r - ∑_{r=1}^{4} u_r.

Solution

We are given the recurrence relation un+1=15unu_{n+1} = \frac{1}{5} u_n and the initial value u1=5u_1 = 5. The problem asks to find the exact value of:

r=1urr=14ur\sum_{r=1}^{\infty} u_r - \sum_{r=1}^{4} u_r

Step 1: Understanding the Sequence

The recurrence relation tells us that each term in the sequence is a fifth of the previous term. So, we can compute the first few terms of the sequence:

  • u1=5u_1 = 5
  • u2=15×u1=15×5=1u_2 = \frac{1}{5} \times u_1 = \frac{1}{5} \times 5 = 1
  • u3=15×u2=15×1=15u_3 = \frac{1}{5} \times u_2 = \frac{1}{5} \times 1 = \frac{1}{5}
  • u4=15×u3=15×15=125u_4 = \frac{1}{5} \times u_3 = \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}
  • u5=15×u4=15×125=1125u_5 = \frac{1}{5} \times u_4 = \frac{1}{5} \times \frac{1}{25} = \frac{1}{125}

Step 2: General Term

From this, we can see that the general term of the sequence is given by:

un=5×(15)n1=5×5(n1)=5×15n1=55n1=15n1.u_n = 5 \times \left( \frac{1}{5} \right)^{n-1} = 5 \times 5^{-(n-1)} = 5 \times \frac{1}{5^{n-1}} = \frac{5}{5^{n-1}} = \frac{1}{5^{n-1}}.

Step 3: Sum of Infinite Series

Now, let's calculate the sum of the infinite series r=1ur\sum_{r=1}^{\infty} u_r. This is a geometric series with the first term u1=5u_1 = 5 and common ratio 15\frac{1}{5}. The sum of an infinite geometric series is given by:

S=a1rS = \frac{a}{1 - r} where aa is the first term and rr is the common ratio.

For this series:

  • a=5a = 5
  • r=15r = \frac{1}{5}

Thus, the sum of the infinite series is:

S=5115=545=5×54=254.S_{\infty} = \frac{5}{1 - \frac{1}{5}} = \frac{5}{\frac{4}{5}} = \frac{5 \times 5}{4} = \frac{25}{4}.

Step 4: Sum of First Four Terms

Next, let's compute the sum of the first four terms:

S4=u1+u2+u3+u4=5+1+15+125.S_4 = u_1 + u_2 + u_3 + u_4 = 5 + 1 + \frac{1}{5} + \frac{1}{25}.

To simplify:

S4=5+1+0.2+0.04=6.24.S_4 = 5 + 1 + 0.2 + 0.04 = 6.24.

Step 5: Final Expression

We need to compute:

r=1urr=14ur=SS4.\sum_{r=1}^{\infty} u_r - \sum_{r=1}^{4} u_r = S_{\infty} - S_4.

Substituting the values:

=2546.24.= \frac{25}{4} - 6.24.

First, let's express 6.246.24 as a fraction:

6.24=624100=15625.6.24 = \frac{624}{100} = \frac{156}{25}.

Now, subtracting the two values:

25415625.\frac{25}{4} - \frac{156}{25}.

We will find a common denominator and subtract the fractions. Let me calculate this step:

LCM of 4 and 25 is 100, so: \text{LCM of 4 and 25 is 100, so: }

254=625100,15625=624100.\frac{25}{4} = \frac{625}{100}, \quad \frac{156}{25} = \frac{624}{100}.

Thus:

625100624100=1100.\frac{625}{100} - \frac{624}{100} = \frac{1}{100}.

Final Answer:

The exact value of the expression is:

1100.\boxed{\frac{1}{100}}.

Would you like more details or have any further questions?


Relative questions:

  1. How do you generalize the sum of geometric series for different common ratios?
  2. What is the effect of different starting values on the infinite sum of the series?
  3. Can the sum of a geometric series ever diverge, and under what conditions?
  4. How do you convert between decimal and fractional representations accurately?
  5. What are the properties of the recurrence relation in the context of this problem?

Tip: Always remember to check the convergence of a series before computing its infinite sum.

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Math Problem Analysis

Mathematical Concepts

Sequences
Recurrence Relations
Geometric Series

Formulas

General term of a geometric sequence: u_n = a * r^(n-1)
Sum of an infinite geometric series: S = a / (1 - r)
Sum of the first n terms of a geometric series

Theorems

Geometric Series Sum Formula

Suitable Grade Level

Grade 10-12 or early university