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Verify that for all \(n \geq 0\), $$ \frac{1}{2}\left(\frac{1}{2 n+1}\right)\left(\begin{array}{c} 2 n+2 \\ n+1 \end{array}\right)=\left(\frac{1}{n+1}\right)\left(\begin{array}{c} 2 n \\ n \end{array}\right) $$

Short Answer

Expert verified
The given identity is correct for all \(n \geq 0\).

Step by step solution

01

Interpret the Given Identity

The given identity involves an equality between two expressions: the left hand side (LHS) and the right hand side (RHS). The expressions have binomial coefficients which are the coefficients of \(x^n\) in the binomial expansion of \((1+x)^n\). Both sides also contain fractions. Now, proceed to simplify both sides in order to verify the given identity.
02

Simplify the Left Hand Side (LHS)

The left hand side (LHS) is \(\frac{1}{2}\) multiplied by \(\frac{1}{2n+1}\) multiplied by the binomial coefficient \(\binom{2n+2}{n+1}\). The binomial coefficient can be expanded into factorial form like so: \(\frac{(2n+2)!}{((n+1)!(2n+2-(n+1))!)}\). After further simplification, the LHS becomes \(\frac{(2n+2)!}{2(n+1)!^2}\). Since \( (2n+2)! = (2n+2)(2n+1)! = 2(n+1)(2n+1)(2n)! \), the LHS simplifies into: \(\frac{2(n+1)(2n+1)(2n)!}{2(n+1)!^2} = \frac{(2n+1)}{2(n+1)}\cdot \frac{(2n)!}{(n!)^2}\).
03

Simplify the Right Hand Side (RHS)

The right hand side (RHS) is \(\frac{1}{n+1}\) multiplied by the binomial coefficient \(\binom{2n}{n}\). The binomial coefficient can be expanded into factorial form like so: \(\frac{(2n)!}{(n!(2n-n)!)}\). After simplification, the RHS becomes \(\frac{(2n)!}{n!(n+1)!}\cdot \frac{1}{n+1} = \frac{(2n!)}{(n!)^2}=\frac{(2n+1)}{2(n+1)}\cdot \frac{(2n)!}{(n!)^2}\).
04

Verify the Identity

Now comparing the simplified LHS and RHS expressions, we notice that they are equal to each other: \(\frac{(2n+1)}{2(n+1)}\cdot \frac{(2n)!}{(n!)^2} = \frac{(2n+1)}{2(n+1)}\cdot \frac{(2n)!}{(n!)^2}\). Thus, the given identity has been verified for all \(n \geq 0\).

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Key Concepts

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

Factorial Notation
Understanding factorial notation is fundamental in tackling problems involving binomial coefficients. The factorial of a non-negative integer \(n\), denoted as \(n!\), is the product of all positive integers less than or equal to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). When we expand expressions like \(\binom{2n+2}{n+1}\), we express it as \(\frac{(2n+2)!}{(n+1)!(2n+1)!}\).
This notation simplifies the manipulation of large numbers and expressions, especially in combinatorial calculations. Factorials also serve to define permutations and combinations, as they provide a means to calculate the number of ways to arrange or select items. It’s essential to note that \(0! = 1\), which can be intriguing but makes sense since any number raised to zero factorial is defined as one.
Combinatorics
Combinatorics is the branch of mathematics concerned with counting and arrangements. It's essential in verifying identities involving binomial coefficients, like the one in our exercise. Binomial coefficients \(\binom{n}{k}\) represent the number of ways to choose \(k\) elements from \(n\) elements without regard to order.
In our given identity, we see expressions such as \(\binom{2n+2}{n+1}\) and \(\binom{2n}{n}\). These coefficients play a crucial role in simplifying and verifying the equality. Knowing how to express these coefficients using factorials, as shown in the solution, allows us to translate complex combinatorial concepts into manageable algebraic forms.
The binomial theorem, which involves coefficients like the ones we deal with, provides insights into how these combinations expand into expressions like \((1+x)^n\). This understanding is essential in calculus and probability, where combinatorics often provides the foundation for deeper insights.
Mathematical Proof
Mathematical proof is the logical process of verifying that a statement is true. In our exercise, the proof involves showing that the left hand side (LHS) and right hand side (RHS) of an equation are equal for all \(n \geq 0\).
The step-by-step simplification process used in our solution is a form of direct proof. By manipulating both sides of the equation using known mathematical identities and simplifications, we demonstrate their equivalence.
This method requires a solid understanding of algebraic manipulations and properties of binomial coefficients. Successfully proving such identities helps establish confidence in mathematical reasoning and problem-solving approaches. Proofs are not just about reaching a conclusion; they are about understanding why something is true, which is an essential skill in all areas of mathematics.

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Most popular questions from this chapter

Determine the number of \(n\)-digit quaternary \((0,1,2,3)\) sequences in which there is never a 3 anywhere to the right of a 0 .

a) For the binary string 001110 , there are three runs: 00 , 111 , and 0 . Meanwhile, the string 000111 has only two runs: 000 and 111; while the string 010101 determines the six runs: \(0,1,0,1,0,1\). For \(n=1\), we consider two binary strings, namely, 0 and 1 - these two strings (of length 1) determine a total of two runs. There are four binary strings of length \(n=2\) and these strings determine 1 (for 00\()+2\) \((\) for 01\()+2(\) for 10\()+1(\) for 11\()=6\) runs. Find and solve a recurrence relation for \(t_{n}\), the total number of runs determined by the \(2^{n}\) binary strings of length \(n\), where \(n \geq 1\). b) Answer the question posed in part (a) for quaternary strings of length \(n\). (Here the alphabet comprises \(0,1,2,3\).) c) Generalize the results of parts (a) and (b).

For \(n \geq 0\), evenly distribute \(2 n\) points on the circumference of a circle, and label these points cyclically with the integers \(1,2,3, \ldots, 2 n\). Let \(a_{n}\) be the number of ways in which these \(2 n\) points can be paired off as \(n\) chords where no two chords intersect. (The case for \(n=3\) is shown in Fig. 10.23.) Find and solve a recurrence relation for \(a_{n}, n \geq 0\).

For \(n \in \mathbf{N}\), let \(s_{n}\) count the number of ways one can travel from \((0,0)\) to \((n, n)\) using the moves \(\mathrm{R}:(x, y) \rightarrow(x+1, y)\), \(\mathrm{U}:(x, y) \rightarrow(x, y+1), \mathrm{D}:(x, y) \rightarrow(x+1, y+1)\), where the path can never rise above the line \(y=x\). (a) Determine \(s_{2}\). (b) How is \(s_{2}\) related to the Catalan numbers \(b_{0}, b_{1}, b_{2} ?\) (c) How is \(s_{3}\) related to \(b_{0}, b_{1}, b_{2}, b_{3} ?\) What is \(s_{3} ?\) (d) For \(n \in \mathbf{N}\), how is \(s_{n}\) related to \(b_{0}, b_{1}, b_{2}, \ldots, b_{n}\) ? (The numbers \(s_{0}, s_{1}, s_{2}, \ldots\) are known as the Schröder numbers.)

For \(n \in \mathbf{Z}^{+}\), let \(f:\\{1,2, \ldots, n\\} \rightarrow\\{1,2, \ldots, n\\}\), where \(f\) is monotone increasing [that is, \(1 \leq i

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