A Sequence Is Defined Recursively Using The Formula $f(n+1) = -0.5 F(n$\]. If The First Term Of The Sequence Is 120, What Is $f(5$\]?A. \[$-15\$\]B. \[$-7.5\$\]C. \[$7.5\$\]D. \[$15\$\]

by ADMIN 186 views

Introduction

In mathematics, recursive sequences are a fundamental concept that has far-reaching implications in various fields, including algebra, calculus, and number theory. A recursive sequence is defined by a formula that specifies how each term is obtained from the previous term. In this article, we will delve into the world of recursive sequences and explore the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n), where the first term of the sequence is 120. Our goal is to find the value of f(5)f(5).

Understanding Recursive Sequences

A recursive sequence is defined by a formula that specifies how each term is obtained from the previous term. The formula is typically of the form f(n+1)=g(f(n))f(n+1) = g(f(n)), where gg is a function that takes the previous term as input and produces the next term. In our case, the formula is f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n), which means that each term is obtained by multiplying the previous term by βˆ’0.5-0.5.

The Formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)

The formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) is a simple yet powerful tool for generating recursive sequences. To understand how it works, let's start with the first term of the sequence, which is 120. We can then use the formula to find the next term, which is f(2)=βˆ’0.5f(1)=βˆ’0.5Γ—120=βˆ’60f(2) = -0.5 f(1) = -0.5 \times 120 = -60. We can continue this process to find the subsequent terms of the sequence.

Finding f(5)f(5)

To find f(5)f(5), we need to apply the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) four times, starting from the first term of the sequence, which is 120. Here's the step-by-step process:

  1. f(2)=βˆ’0.5f(1)=βˆ’0.5Γ—120=βˆ’60f(2) = -0.5 f(1) = -0.5 \times 120 = -60
  2. f(3)=βˆ’0.5f(2)=βˆ’0.5Γ—(βˆ’60)=30f(3) = -0.5 f(2) = -0.5 \times (-60) = 30
  3. f(4)=βˆ’0.5f(3)=βˆ’0.5Γ—30=βˆ’15f(4) = -0.5 f(3) = -0.5 \times 30 = -15
  4. f(5)=βˆ’0.5f(4)=βˆ’0.5Γ—(βˆ’15)=7.5f(5) = -0.5 f(4) = -0.5 \times (-15) = 7.5

Conclusion

In this article, we explored the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) and used it to find the value of f(5)f(5), given that the first term of the sequence is 120. We saw that the formula is a simple yet powerful tool for generating recursive sequences and that it can be used to find the value of any term in the sequence. The value of f(5)f(5) is 7.5.

Answer

The correct answer is C. f(5)=7.5f(5) = 7.5.

Discussion

This problem is a great example of how recursive sequences can be used to model real-world phenomena. In this case, the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) can be used to model a situation where a quantity is decreasing by half at each step. This type of problem is commonly encountered in fields such as finance, economics, and engineering.

Related Problems

  • Find the value of f(10)f(10), given that the first term of the sequence is 120 and the formula is f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n).
  • Find the value of f(5)f(5), given that the first term of the sequence is 240 and the formula is f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n).
  • Find the value of f(10)f(10), given that the first term of the sequence is 120 and the formula is f(n+1)=2f(n)f(n+1) = 2 f(n).

Solutions

  • f(10)=βˆ’0.59Γ—120=βˆ’0.59Γ—120=βˆ’0.001953125Γ—120=βˆ’0.23534375f(10) = -0.5^{9} \times 120 = -0.5^{9} \times 120 = -0.001953125 \times 120 = -0.23534375
  • f(5)=βˆ’0.54Γ—240=βˆ’0.54Γ—240=βˆ’0.0625Γ—240=βˆ’15f(5) = -0.5^{4} \times 240 = -0.5^{4} \times 240 = -0.0625 \times 240 = -15
  • f(10)=29Γ—120=29Γ—120=512Γ—120=61440f(10) = 2^{9} \times 120 = 2^{9} \times 120 = 512 \times 120 = 61440
    A Recursive Sequence: Unraveling the Mystery of f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) ===========================================================

Q&A: A Recursive Sequence

Q: What is a recursive sequence?

A: A recursive sequence is a sequence of numbers that is defined by a formula that specifies how each term is obtained from the previous term.

Q: What is the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)?

A: The formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) is a recursive formula that specifies how each term in the sequence is obtained from the previous term. It means that each term is obtained by multiplying the previous term by βˆ’0.5-0.5.

Q: How do I find the value of f(5)f(5) using the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)?

A: To find the value of f(5)f(5), you need to apply the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) four times, starting from the first term of the sequence, which is 120. Here's the step-by-step process:

  1. f(2)=βˆ’0.5f(1)=βˆ’0.5Γ—120=βˆ’60f(2) = -0.5 f(1) = -0.5 \times 120 = -60
  2. f(3)=βˆ’0.5f(2)=βˆ’0.5Γ—(βˆ’60)=30f(3) = -0.5 f(2) = -0.5 \times (-60) = 30
  3. f(4)=βˆ’0.5f(3)=βˆ’0.5Γ—30=βˆ’15f(4) = -0.5 f(3) = -0.5 \times 30 = -15
  4. f(5)=βˆ’0.5f(4)=βˆ’0.5Γ—(βˆ’15)=7.5f(5) = -0.5 f(4) = -0.5 \times (-15) = 7.5

Q: What is the value of f(10)f(10) using the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)?

A: To find the value of f(10)f(10), you need to apply the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) nine times, starting from the first term of the sequence, which is 120. Here's the step-by-step process:

  1. f(2)=βˆ’0.5f(1)=βˆ’0.5Γ—120=βˆ’60f(2) = -0.5 f(1) = -0.5 \times 120 = -60
  2. f(3)=βˆ’0.5f(2)=βˆ’0.5Γ—(βˆ’60)=30f(3) = -0.5 f(2) = -0.5 \times (-60) = 30
  3. f(4)=βˆ’0.5f(3)=βˆ’0.5Γ—30=βˆ’15f(4) = -0.5 f(3) = -0.5 \times 30 = -15
  4. f(5)=βˆ’0.5f(4)=βˆ’0.5Γ—(βˆ’15)=7.5f(5) = -0.5 f(4) = -0.5 \times (-15) = 7.5
  5. f(6)=βˆ’0.5f(5)=βˆ’0.5Γ—7.5=βˆ’3.75f(6) = -0.5 f(5) = -0.5 \times 7.5 = -3.75
  6. f(7)=βˆ’0.5f(6)=βˆ’0.5Γ—(βˆ’3.75)=1.875f(7) = -0.5 f(6) = -0.5 \times (-3.75) = 1.875
  7. f(8)=βˆ’0.5f(7)=βˆ’0.5Γ—1.875=βˆ’0.9375f(8) = -0.5 f(7) = -0.5 \times 1.875 = -0.9375
  8. f(9)=βˆ’0.5f(8)=βˆ’0.5Γ—(βˆ’0.9375)=0.46875f(9) = -0.5 f(8) = -0.5 \times (-0.9375) = 0.46875
  9. f(10)=βˆ’0.5f(9)=βˆ’0.5Γ—0.46875=βˆ’0.234375f(10) = -0.5 f(9) = -0.5 \times 0.46875 = -0.234375

Q: How do I find the value of f(n)f(n) using the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)?

A: To find the value of f(n)f(n), you need to apply the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) nβˆ’1n-1 times, starting from the first term of the sequence, which is 120. Here's the general step-by-step process:

  1. f(2)=βˆ’0.5f(1)=βˆ’0.5Γ—120=βˆ’60f(2) = -0.5 f(1) = -0.5 \times 120 = -60
  2. f(3)=βˆ’0.5f(2)=βˆ’0.5Γ—(βˆ’60)=30f(3) = -0.5 f(2) = -0.5 \times (-60) = 30
  3. f(4)=βˆ’0.5f(3)=βˆ’0.5Γ—30=βˆ’15f(4) = -0.5 f(3) = -0.5 \times 30 = -15
  4. ... nβˆ’1n-1 steps later... nn steps later...

Q: What is the general formula for f(n)f(n) using the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n)?

A: The general formula for f(n)f(n) is:

f(n)=(βˆ’0.5)nβˆ’1Γ—120f(n) = (-0.5)^{n-1} \times 120

This formula can be used to find the value of f(n)f(n) for any positive integer nn.

Conclusion

In this article, we explored the formula f(n+1)=βˆ’0.5f(n)f(n+1) = -0.5 f(n) and used it to find the value of f(5)f(5) and f(10)f(10). We also provided a general formula for f(n)f(n) and explained how to use it to find the value of f(n)f(n) for any positive integer nn.