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Find \(P_{k+1}\) for the given \(P_{k}\). $$P_{k}=\frac{k^{2}(k+3)^{2}}{6}$$

Short Answer

Expert verified
The term \(P_{k+1}\) in the sequence, based on the rule \(P_{k}=\frac{k^{2}(k+3)^{2}}{6}\), is \(P_{k+1}=\frac{(k+1)^{2}(k+4)^{2}}{6}\)

Step by step solution

01

Replace \(k\) with \(k+1\) in the rule for \(P_{k}\)

Replace every instance of \(k\) in the rule \(P_{k}=\frac{k^{2}(k+3)^{2}}{6}\) with \(k+1\) to get: \(P_{k+1}=\frac{(k+1)^{2}((k+1)+3)^{2}}{6}\).
02

Simplify \((k+1)+3\)

Simplify the \((k+1)+3\) in the numerator of \(P_{k+1}\) from the previous step to get \(P_{k+1}=\frac{(k+1)^{2}(k+4)^{2}}{6}\).
03

Expand and Simplify

Expand and simplify the expression \(P_{k+1}=\frac{(k+1)^{2}(k+4)^{2}}{6} = \frac{(k^2 + 2k + 1)(k^2 + 8k +16)}{6}\)

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Key Concepts

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

Algebraic Expressions
Algebraic expressions are mathematical phrases that can contain numbers, variables, and operators. They form the backbone of many mathematical concepts, offering a way to generalize numbers and operations. These expressions can represent a variety of scenarios, such as the fundamental relationships between variables. For example, given the expression \(P_k = \frac{k^2(k+3)^2}{6}\), we see a blend of variables and constants forming a complex expression commonly used in problems involving sequences and series.

Key components of algebraic expressions include:
  • Terms: These are the building blocks of expressions. Each term can be a number, a variable, or the product of both. In our example, \(k^2\) and \((k+3)^2\) are terms.
  • Operators: Symbols such as addition and multiplication connect terms. Here, multiplication combines the terms \((k^2)\) and \((k+3)^2\).
  • Coefficients: These are the numerical parts of the terms, such as the 1 (implied) in front of \(k^2\).
Algebraic expressions can often be simplified to make them more manageable. This includes combining like terms or factoring, as seen in the process of finding \(P_{k+1}\) from \(P_k\) by replacing \(k\) with \(k+1\) and simplifying.
Polynomial Expansion
Polynomial expansion is a crucial technique used to simplify expressions that involve powers of terms in a bracket. This technique eliminates brackets by spreading or distributing the multiplication over addition. It leverages the distributive property, which states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products.

Consider the expression \((k+1)^2(k+4)^2\) in our problem. When expanding, each term in the first polynomial multiplies with each term in the second polynomial. This results in a new polynomial:
  • First, expand \((k+1)^2\) to get \(k^2 + 2k + 1\).
  • Then, expand \((k+4)^2\) to get \(k^2 + 8k + 16\).
  • Multiply both expanded forms to obtain the expression \((k^2 + 2k + 1)(k^2 + 8k + 16)\).
The expanded form results in a more complex polynomial where each term is simplified further by multiplying and combining like terms. In our step-by-step process, this allowed us to transition from \((k+1)^2(k+4)^2\) directly to a fully expanded and simplified polynomial.
Indices and Exponents
Indices and exponents are mathematical notations used to represent numbers raised to a specific power. Understanding this concept is essential to manipulating expressions involving powers, such as polynomials and sequences. The base number is the number being multiplied, while the exponent is the power indicating how many times the base is used in the multiplication.

In the case of \((k+1)^2\), for example, 2 is the exponent, indicating that \(k+1\) is multiplied by itself. The process of expanding this involves applying this repeated multiplication:
  • Squaring: For \((k+1)^2\), you calculate \((k+1)(k+1)\), which results in \(k^2 + 2k + 1\).
  • For another term, \((k+4)^2\), similarly compute \((k+4)(k+4)\) resulting in \(k^2 + 8k + 16\).
Understanding how indices work is pivotal in handling more complex situations that involve higher powers or roots. They simplify calculations in algebraic expressions and polynomial expansions, making solutions more systematic and less prone to errors. This methodical approach helps efficiently handle the transition from \(P_k\) to \(P_{k+1}\).

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Most popular questions from this chapter

Use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \(\begin{aligned}(1.02)^{8} &=(1+0.02)^{8} \\ &=1+8(0.02)+28(0.02)^{2}+\cdot \cdot \cdot+(0.02)^{8}\end{aligned}\), $$(2.005)^{10}$$

The table shows the average prices \(f(t)\) (in cents per kilowatt hour) of residential electricity in the United States from 2003 through 2010 . (Source: U.S. Energy Information Administration ).$$\begin{array}{|c|c|}\hline \text { Year } & \text { Abcrage Poids }(10) \\\\\hline 2003 & 8.72 \\\2004 & 8.95 \\\2005 & 9.45 \\\2006 & 10.40 \\\2007 & 10.65 \\\2008 & 11.26 \\\2009 & 11.51 \\\2010 & 11.58 \\\\\hline\end{array}$$.(a) Use the regression feature of a graphing utility to find a cubic model for the data. Let \(t\) represent the year, with \(t=3\) corresponding to 2003 (b) Use the graphing utility to plot the data and the model in the same viewing window. (c) You want to adjust the model so that \(t=3\) corresponds to 2008 rather than \(2003 .\) To do this, you shift the graph of \(f\) five units to the left to obtain \(g(t)=f(t+5) .\) Use binomial coefficients to write \(g(t)\) in standard form. (d) Use the graphing utility to graph \(g\) in the same viewing window as \(f\) (e) Use both models to predict the average price in \(2011 .\) Do you obtain the same answer? (f) Do your answers to part (e) seem reasonable? Explain. (g) What factors do you think may have contributed to the change in the average price?

Determine whether the statement is true or false. Justify your answer. $$\sum_{i=1}^{4}\left(i^{2}+2 i\right)=\sum_{i=1}^{4} i^{2}+2 \sum_{i=1}^{4} i$$

Consider \(n\) independent trials of an experiment in which each trial has two possible outcomes: "success" or "failure." The probability of a success on each trial is \(p,\) and the probability of a failure is \(q=1-p .\) In this context, the term \(_{n} C_{k} p^{k} q^{n-k}\) in the expansion of \((p+q)^{n}\) gives the probability of \(k\) successes in the \(n\) trials of the experiment.To find the probability that the sales representative in Exercise 87 makes four sales when the probability of a sale with any one customer is \(\frac{1}{2},\) evaluate the term $$_{8} C_{4}\left(\frac{1}{2}\right)^{4}\left(\frac{1}{2}\right)^{4}$$, in the expansion of \(\left(\frac{1}{2}+\frac{1}{2}\right)^{8}\).

Complete each expression for the apparent \(n\) th term \(a_{n}\) of the sequence. Which expressions are appropriate to represent the cost \(a_{n}\) to buy \(n\) MP3 songs at a cost of \(\$ 1\) per song? Explain. $$\text { (a) } a_{n}=1 \square$$ $$\text { (b) } a_{n}=\frac{ \square 1}{(n-1) !}$$ $$\text { (c) } a_{n}=\sum_{k=1}^{n}$

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