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Finding \(P_{k+1}\) Given \(P_{k}\) In Exercises \(5-10,\) find \(P_{k+1}\) for the given \(P_{k} .\) $$P_{k}=\frac{k^{2}}{2(k+1)^{2}}$$

Short Answer

Expert verified
The value of \(P_{k+1}\) based on the given \(P_k\) is \(P_{k + 1} = \frac{(k+1)^{2}}{2(k+2)^{2}}\).

Step by step solution

01

Understand the Question

Understand that we're trying to find the next term in the sequence, \(P_{k+1}\), which involves replacing \(k\) with \(k+1\) in the given equation for \(P_k\).
02

Substitute \(k+1\) for \(k\)

Replace \(k\) with \(k + 1\) in the given equation \(P_{k}=\frac{k^{2}}{2(k+1)^{2}}\). This results in \(P_{k + 1}=\frac{(k+1)^{2}}{2((k+1)+1)^{2}}\).
03

Simplify the Equation

Simplify this equation as much as possible. Here, we could simplify the denominator as \((k+2)^{2}\): \(P_{k + 1}=\frac{(k+1)^{2}}{2(k+2)^{2}}\).

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Key Concepts

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

Arithmetic Sequences
Arithmetic sequences are a staple in mathematics, fundamental for understanding how patterns in numbers develop. An arithmetic sequence is essentially a list of numbers where each term after the first is created by adding a constant to the previous term. This constant is known as the common difference, usually denoted by 'd'.

Consider the simple arithmetic sequence 2, 4, 6, 8, ... Here, each term is 2 more than the term before it, making the common difference 2. This is written mathematically as:
\[a_n = a_1 + (n-1)d\]
Where \(a_n\) is the nth term of the sequence, \(a_1\) is the first term, and 'n' represents the term number. By knowing the initial term and the common difference, we can calculate any term in the sequence, which becomes particularly useful in solving problems related to linear growth or decline.
Geometric Sequences
In contrast to arithmetic sequences, geometric sequences grow by a factor, not by addition. A geometric sequence is an ordered list of numbers where each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio ('r').

For example, the sequence 3, 6, 12, 24, ... is geometric because each term is twice as large as the term before it hence, the common ratio is 2. The general form of a geometric sequence can be expressed as:
\[a_n = a_1 \times r^{(n-1)}\]
Where \(a_n\) is the nth term, \(a_1\) is the first term, and 'n' is the term position within the sequence. Geometric sequences are extremely useful when dealing with exponential growth scenarios, such as population growth, interest rates, and certain natural phenomena.
Recursive Formulas
Recursive formulas provide another angle to defining sequences, especially when a direct formula is difficult to ascertain. These formulas take the previous terms of a sequence to define the next one. In essence, each term is based on its predecessors, offering a step-by-step approach to sequence construction.

A recursive formula has two parts: the starting value, which is the initial term or terms of the sequence, and the recursion equation, which tells you how to get the term from its preceding term(s). For arithmetic and geometric sequences, a recursive formula is usually straightforward. In the case of arithmetic, it looks like this:
\[a_{n} = a_{n-1} + d\]
For geometric sequences, it takes the form:
\[a_{n} = a_{n-1} \times r\]
However, for more complex sequences, the recursion might involve several past terms or even the position of the term itself, as is illustrated by the problem defining \(P_{k+1}\) with \(P_k\). Recursive formulas are immensely valuable in computer science for writing algorithms and in mathematics for summing series or solving complex numerical patterns where other formulas are impractical.

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Most popular questions from this chapter

Proof In Exercises \(99-102,\) prove the property for all integers \(r\) and \(n\) where \(0 \leq r \leq n .\) $$_{n+1} C_{r}=_{n} C_{r}+_{n} C_{r-1}$$

In Exercises \(75-82,\) solve for \(n\) $$_{n} P_{6}=12 \cdot_{n-1} P_{5}$$

Consider a group of \(n\) people. (a) Explain why the following pattern gives the probabilities that the \(n\) people have distinct birthdays. $$\begin{array}{l}{n=2 : \frac{365}{365} \cdot \frac{364}{365}=\frac{365 \cdot 364}{365^{2}}} \\ {n=3 : \frac{365}{365} \cdot \frac{364}{365} \cdot \frac{363}{365}=\frac{365 \cdot 364 \cdot 363}{365^{3}}}\end{array}$$ (b) Use the pattern in part (a) to write an expression for the probability that \(n=4\) people have distinct birthdays. (c) Let \(P_{n}\) be the probability that the \(n\) people have distinct birthdays. Verify that this probability can be obtained recursively by $$P_{1}=1\( and \)P_{n}=\frac{365-(n-1)}{365} P_{n-1}$$ (d) Explain why \(Q_{n}=1-P_{n}\) gives the probability that at least two people in a group of \(n\) people have the same birthday. (e) Use the results of parts (c) and (d) to complete the table. (f) How many people must be in a group so that the probability of at least two of them having the same birthday is greater than \(\frac{1}{2} ?\) Explain.

Finding a Sum In Exercises \(45-54\) , find the sum using the formulas for the sums of powers of integers. $$\sum_{n=1}^{30} n$$

The sample spaces are large and you should use the counting principles discussed in Section 9.6. ATM personal identification number (PIN) codes typically consist of four-digit sequences of numbers. Find the probability that if you forget your PIN, then you can guess the correct sequence (a) at random and (b) when you recall the first two digits.

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