Math Problem Statement

If f, of, 1, equals, 3f(1)=3 and f, of, n, plus, 1, equals, f, of, n, squared, plus, 1f(n+1)=f(n)2+1 then find the value of f, of, 4f(4)

Solution

Let's break down the problem systematically. The function f(n)f(n) is defined by the following:

  • f(1)=3f(1) = 3
  • f(n+1)=f(n)2+1f(n+1) = f(n)^2 + 1

We need to find the value of f(4)f(4).

Step 1: Calculate f(2)f(2)

Using the recurrence relation: f(2)=f(1)2+1=32+1=9+1=10f(2) = f(1)^2 + 1 = 3^2 + 1 = 9 + 1 = 10

Step 2: Calculate f(3)f(3)

f(3)=f(2)2+1=102+1=100+1=101f(3) = f(2)^2 + 1 = 10^2 + 1 = 100 + 1 = 101

Step 3: Calculate f(4)f(4)

f(4)=f(3)2+1=1012+1f(4) = f(3)^2 + 1 = 101^2 + 1

Now, calculate 1012101^2: 1012=10201101^2 = 10201 So: f(4)=10201+1=10202f(4) = 10201 + 1 = 10202

Final Answer:

The value of f(4)f(4) is 10202\boxed{10202}.

Would you like more details or have any questions? Here are some related questions:

  1. What pattern do you observe in the values of f(n)f(n)?
  2. How would the values change if f(1)f(1) were different?
  3. Can you generalize a formula for f(n)f(n)?
  4. What would f(5)f(5) be?
  5. How does the recurrence relation affect the growth of f(n)f(n)?

Tip: In recurrence relations, always calculate a few initial terms to understand the pattern and behavior of the function.

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Math Problem Analysis

Mathematical Concepts

Recurrence Relations
Sequence Analysis

Formulas

Recurrence relation: f(n+1) = f(n)^2 + 1

Theorems

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Suitable Grade Level

Grades 10-12