Show that if \(p\) is a positive integer, then \(\left(\begin{array}{l}p \\\ n\end{array}\right)=0\) when \(n>p,\) so \((1+x)^{p}=\sum\left(\begin{array}{l}p \\\ n\end{array}\right) x^{n}\) is just a sum of \(p+1\) terms, from \(n=0\) to \(n=p .\) For example, \((1+x)^{2}\) has 3 terms, \((1+x)^{3}\) has 4 terms, etc. This is just the familiar binomial theorem.

Short Answer

Expert verified
If \( n > p \), \ (\binom{p}{n} = 0 \. Thus, \( (1 + x)^p \) only has \ p + 1 \ terms from \( n = 0 \) to \( n = p \).

Step by step solution

01

Understand the Binomial Coefficient

The binomial coefficient \(\binom{p}{n}\) is defined as \(\frac{p!}{n!(p-n)!}\). This coefficient represents the number of ways to choose \ n \ elements from a set of \ p \ elements.
02

Condition When \( n > p \)

When \( n \) is greater than \( p \), the term \( (p - n)! \) in the denominator becomes negative, leading to an undefined factorial. Specifically, for \( n > p \), \( \binom{p}{n} = \frac{p!}{n!(p-n)!} = 0 \). Hence, this term does not contribute to the sum.
03

Formulating the Binomial Expansion

For binomials, we use \( (1 + x)^p = \binom{p}{0} x^0 + \binom{p}{1} x^1 + \binom{p}{2} x^2 + \ \dots \ \binom{p}{p} x^p \). Given from Step 2, when \( n > p \), \(\binom{p}{n}\) becomes 0, so the series stops at \( n = p \). Therefore, this expansion contains \ p+1 \ terms (from \( n = 0 \) to \( n = p \)).
04

Conclude with Examples

To illustrate, \( (1 + x)^2 = \binom{2}{0} x^0 + \binom{2}{1} x^1 + \binom{2}{2} x^2 \), and includes 3 terms. Similarly, \( (1 + x)^3 = \binom{3}{0} x^0 + \binom{3}{1} x^1 + \binom{3}{2} x^2 + \binom{3}{3} x^3 \) includes 4 terms.

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Key Concepts

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

Binomial Coefficient
The binomial coefficient is a key concept in combinatorics. It is denoted as \(\binom{p}{n}\) and is defined mathematically as: \(\binom{p}{n} = \frac{p!}{n!(p-n)!}\). This represents the number of ways to choose \(\text{n}\) elements from a set of \(\text{p}\) elements.
Factors in the formula include factorials. A factorial, denoted by \(!\), is the product of all positive integers less than or equal to that number. For instance:
  • \(4! = 4 \times 3 \times 2 \times 1 = 24\)
  • \(3! = 3 \times 2 \times 1 = 6\)
It's important to note that when \(\text{n} > \text{p}\), the term \( (p - n)! \) becomes negative. This results in the binomial coefficient being zero because factorials for negative values are undefined. Hence, \(\binom{p}{n} = 0\) if \(\text{n} > \text{p}\).
Understanding this property is crucial, especially when dealing with binomial expansions.
Binomial Expansion
The binomial theorem allows us to expand expressions of the form \((1 + x)^p\) into a sum of terms involving binomial coefficients. The general formula is given by: \((1 + x)^p = \binom{p}{0} x^0 + \binom{p}{1} x^1 + \binom{p}{2} x^2 + ... + \binom{p}{p} x^p\).
This formula signifies that each term in the expansion is a product of a binomial coefficient and a power of \(x\). For example:
  • For \((1 + x)^2\), the expanded form is \(\binom{2}{0} x^0 + \binom{2}{1} x^1 + \binom{2}{2} x^2 = 1 + 2x + x^2\)
  • For \((1 + x)^3\), the expanded form is \(\binom{3}{0} x^0 + \binom{3}{1} x^1 + \binom{3}{2} x^2 + \binom{3}{3} x^3 = 1 + 3x + 3x^2 + x^3\)

From the previous section, we know that the binomial coefficient \( \binom{p}{n} = 0 \) when \( n > p \). Hence, the series has exactly \(\text{p} + 1\) terms, starting from \(\text{n} = 0\) to \(\text{n} = p\). This is crucial for simplifying calculations and identifying patterns in series.
Factorial
Factorials play an essential role in mathematical calculations and combinatorics. Factorials are denoted by an exclamation mark, like \( n! \), and are defined as the product of all positive integers less than or equal to \(n\). Here are some key points about factorials:
  • \(0! = 1\) by definition.
  • \(n! = n \times (n-1) \times (n-2) \times ... \times 1\)
  • Factorials grow very quickly. For instance, \(5! = 120\) and \(10! = 3,628,800\).

In the context of the binomial coefficient, factorials help determine the different ways to choose subsets of elements. The formula for the binomial coefficient involves dividing the factorial of \(p\) by the product of factorials of \(n\) and \(p-n\). Factorials also appear in probability and statistics, permutations, and many fields of mathematical study.

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Most popular questions from this chapter

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\frac{e^{x}}{1-x}$$

There are 9 one-digit numbers ( 1 to 9 ), 90 two-digit numbers ( 10 to 99 ). How many three-digit, four-digit, etc., numbers are there? The first 9 terms of the harmonic series \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\) are all greater than \(\frac{1}{10}\); similarly consider the next 90 terms, and so on. Thus prove the divergence of the harmonic series by comparison with the series \(\left[\frac{1}{10}+\frac{1}{10}+\cdots\left(9 \text { terms each }=\frac{1}{10}\right)\right]+\left[90 \text { terms each }=\frac{1}{100}\right]+\cdots\) $$ =\frac{9}{10}+\frac{90}{100}+\cdots=\frac{9}{10}+\frac{9}{10}+\cdots $$ The comparison test is really the basic test from which other tests are derived. It is probably the most useful test of all for the experienced mathematician but it is often hard to think of a satisfactory \(m\) series until you have had a good deal of experience with series. Consequently, you will probably not use it as often as the next three tests.

Find the Maclaurin series for the following functions. $$\cos [\ln (1+x)]$$

Use Maclaurin series to evaluate each of the following. Although you could do them by computer, you can probably do them in your head faster than you can type them into the computer. So use these to practice quick and skillful use of basic series to make simple calculations. $$\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{x^{2}}$$

Show that \(\sum_{n=2}^{\infty} 1 / n^{3 / 2}\) is convergent. What is wrong with the following "proof" that it diverges? $$\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{27}}+\frac{1}{\sqrt{64}}+\frac{1}{\sqrt{125}}+\cdots>\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{81}}+\frac{1}{\sqrt{144}}+\cdots$$ which is $$\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\dots=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\right)$$. since the harmonic series diverges, the original series diverges. Hint: Compare \(3 n\) and \(n \sqrt{n}\).

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