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A certain bacterium propagates itself by subdividing, creating four additional bacteria, each identical to the parent bacterium. If the bacteria subdivide in this manner \(n\) times, then, assuming that none of the bacteria die, the number of bacteria present after each subdivision is given by the sequence \(\left\\{B_{k}\right\\}_{k=0}^{n}\), where $$ B_{k}=\frac{4^{k+1}-1}{3} $$

Short Answer

Expert verified
Use the formula \( B_k = \frac{4^{k+1} - 1}{3} \) to find the number of bacteria after each subdivision.

Step by step solution

01

Understanding the Problem

The problem involves calculating the number of bacteria present after each subdivision. The bacteria split into four new bacteria, and they continue to subdivide in the same pattern. The formula given, \( B_k = \frac{4^{k+1} - 1}{3} \), helps in calculating the number of bacteria at the \(k\)-th stage of subdivision.
02

Initial Subdivision (k=0)

For \(k=0\), substitute \(k\) into the formula: \( B_0 = \frac{4^{0+1} - 1}{3} = \frac{4^1 - 1}{3} = \frac{4 - 1}{3} = 1 \). Initially, there's only one bacterium, which is the parent bacterium.
03

Subsequent Subdivisions

Calculate for \(k=1\) by substituting \(k\): \( B_1 = \frac{4^{1+1} - 1}{3} = \frac{4^2 - 1}{3} = \frac{16 - 1}{3} = 5 \). After the first subdivision, there are 5 bacteria.
04

Continuing the Pattern

Calculate for \(k=2\) using the formula: \( B_2 = \frac{4^{2+1} - 1}{3} = \frac{4^3 - 1}{3} = \frac{64 - 1}{3} = 21 \). Two subdivisions result in 21 bacteria.
05

Generalization for nth Subdivision

For any \(k\), use the given formula. It ensures the growth pattern is consistent. As \(k\) increases, the number increases exponentially.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bacterial Propagation
Bacterial propagation is a fascinating process where bacteria reproduce rapidly, leading to exponential growth under ideal conditions. In this exercise, each bacterium divides to produce four new bacteria that are identical to the parent. This process of subdivision occurs without any of the bacteria dying, illustrating a perfect propagation scenario.
Bacteria propagation occurs in discrete, predictable stages. Each stage of subdivision represents another level of bacteria doubling, which in this scenario, quadrupling. With each division, the bacterial population increases by a significant amount, showcasing the efficient nature of bacterial reproduction.
Sequence Formula
The sequence formula given in the exercise is essential for determining the number of bacteria at each stage of subdivision. For this exercise, the sequence formula is given as: \( B_k = \frac{4^{k+1} - 1}{3} \).
Understanding this formula is crucial, as it incorporates both an exponential function and a constant adjustment factor. The exponential term, \(4^{k+1}\), represents the rapid increase in the number of bacteria controlled by the base of the exponent (4) which corresponds to producing four new bacteria per division.
The subtraction of 1 and division by 3 in the formula adjust the value obtained from the exponential term to fit the actual number of bacteria at each stage, ensuring the count starts accurately from the initial single bacterium.
Subdivision Process
The subdivision process is the method by which bacteria multiply in this scenario. When each bacterium divides, it results in four new bacteria. This process can be modeled mathematically by the given formula.
The formula reveals that as bacteria subdivide, the population grows at an exponential rate. This subdivision process is essential for understanding the full impact of bacterial reproduction in scenarios such as population growth in biology or contamination in health sciences.
  • At \(k=0\), the bacterium count starts at 1, representing the original bacterium.
  • For \(k=1\), we calculate using the formula, resulting in 5 bacteria. This includes the original bacterium and its four new descendants.
  • At \(k=2\), the number of bacteria increases further to 21, showing the effect of this rapid multiplication process.
The subdivision process thus illustrates a powerful and dramatic increase in the bacterial population, leading to large numbers in just a few cycles.
Growth Pattern Analysis
Growth pattern analysis helps us understand how the bacterial population evolves over time, following the exponential growth model. In the scenario provided, we observe that each division cycle does not just add to the population but multiplies it significantly.
With the growth pattern described by the formula \( B_k = \frac{4^{k+1} - 1}{3} \), each stage is a clear demonstration of exponential growth. The liver value (4) reflects the robustness of this increase.
By analyzing this pattern, we see the rapid escalation in numbers, showing how a small number of bacteria can soon dominate an environment if left unchecked. This type of analysis is critical in various fields, including microbiology, epidemiology, and environmental sciences, to predict bacterial behavior and implement control measures effectively.

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Most popular questions from this chapter

Two possible methods of adding five numbers are $$ (((a+b)+c)+d)+e $$ and $$ (((e+d)+c)+b)+a $$ Using 4 dp floating-point arithmetic, evaluate the sum $$ \begin{aligned} 10^{1}(0.1000) &+10^{1}(0.1000)-10^{9}(0.5000) \\ &+10^{0}(0.1667)+10^{-1}(0.4167) \end{aligned} $$by both methods. Explain any discrepancy in the results.

Factorize the following: (a) \(a x-2 x-a+2\) (b) \(a^{2}-b^{2}+2 b c-c^{2}\) (c) \(4 k^{2}+4 k l+l^{2}-9 m^{2}\) (d) \(p^{2}-3 p q+2 q^{2}\) (e) \(l^{2}+l m+I n+m n\)

Using exact arithmetic, compute the values of the expressions below. Assuming that all the numbers given are correctly rounded, find absolute and relative error bounds for each term in the expressions and for your answers. Give the answers as correctly rounded numbers. (a) \(1.316-5.713+8.010\) (b) \(2.51 \times 1.01\) (c) \(19.61+21.53-18.67\)

Determine the error bound and relative error bound for \(x\), where (a) \(x=35 \mathrm{~min} \pm 5 \mathrm{~s}\) (b) \(x=35 \mathrm{~min} \pm 4 \%\) (c) \(x=0.58\) and \(x\) is correctly rounded to \(2 \mathrm{dp}\).

A family of straight lines in the \(x-y\) plane is such that each line joins the point \((-p, p)\) on the line \(y=-x\) to the point \((10-p, 10-p)\) on the line \(y=x\), as shown in Figure \(1.38\), for different values of \(p .\) On a piece of graph paper, draw the lines corresponding to \(p=1,2,3, \ldots, 9 .\) The resulting family is seen to envelop a curve. Show that the line which joins \((-p, p)\) to \((10-p,\), \(10-p\) ) has equation $$ 5 y=5 x-p x+10 p-p^{2} $$ Show that two lines of the family pass through the point \(\left(x_{0}, y_{0}\right)\) if \(x_{0}^{2}>20\left(y_{6}-5\right)\), but no lines pass through \(\left(x_{0}, y_{0}\right)\) if \(x_{0}^{2}<20\left(y_{0}-5\right)\). Deduce that the enveloping curve of the family of straight lines is $$ y=\frac{1}{20} x^{2}+5 $$

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