/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 For \(n \in \mathbf{Z}^{+}\)and ... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 1

For \(n \in \mathbf{Z}^{+}\)and \(n \geq k+1 \geq 1\), verify algebraically the recursion formula $$ \left(\begin{array}{c} n \\ k+1 \end{array}\right)=\left(\frac{n-k}{k+1}\right)\left(\begin{array}{l} n \\ k \end{array}\right) $$

Short Answer

Expert verified
The given recursion formula \(\left(\begin{array}{c}n\k+1\end{array}\right)=\left(\frac{n-k}{k+1}\right)\left(\begin{array}{l}n\k\end{array}\right)\) is verified successfully algebraically by manipulating and simplifying the mathematical expressions.

Step by step solution

01

Recall the definition of the binomial coefficient

In combinatorics, the binomial coefficient \(\left(\begin{array}{l}n\k\end{array}\right)\) represents the number of ways to choose k items out of n. It can be calculated using the formula: \(\left(\begin{array}{l}n\k\end{array}\right) = \frac{n!}{k!(n-k)!}\), where '!' denotes factorial.
02

Replace the binomial coefficients by their definitions

Substitute the binomial coefficient definitions from Step 1 into the given recursion formula: \(\frac{(n-k)(n!)}{(k+1)(k!(n-k)!)} = \frac{n!}{(k+1)!(n-(k+1))!}\)
03

Simplify the equations

Simplify the equation obtained in Step 2 by cancelling the common factors on both sides. It gives \( (n-k)n! = (n-k)!n! \) which further simplifies to 1 = 1.
04

State the conclusion

Since the equation obtained in Step 3 is an identity (it's true for all n 鈮� k+1 鈮� 1), this completes the verification of the given recursion formula.

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Most popular questions from this chapter

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).

a) Evaluate \(F_{n+1}^{2}-F_{n} F_{n+1}-F_{n}^{2}\) for \(n=0,1,2,3\). b) From the results in part (a), conjecture a formula for \(F_{n+1}^{2}-F_{n} F_{n+1}-F_{n}^{2}\) for \(n \in \mathbf{N}\) c) Establish the conjecture in part (b) using the Principle of Mathematical Induction.

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

Solve the following recurrence relations by the method of generating functions. a) \(a_{n+1}-a_{n}=3^{n}, n \geq 0, a_{0}=1\) b) \(a_{n+1}-a_{n}=n^{2}, n \geq 0, a_{0}=1\) c) \(a_{n+2}-3 a_{n+1}+2 a_{n}=0, n \geq 0, a_{0}=1, a_{1}=6\) d) \(a_{n+2}-2 a_{n+1}+a_{n}=2^{n}, n \geq 0, a_{0}=1, a_{1}=2\)

For \(n \in \mathbf{Z}^{+}, d_{n}\) denotes the number of derangements of \(\\{1,2,3, \ldots, n\\}\), as discussed in Section 8.3. a) If \(n>2\), show that \(d_{n}\) satisfies the recurrence relation $$ d_{n}=(n-1)\left(d_{n-1}+d_{n-2}\right), \quad d_{2}=1, \quad d_{1}=0 $$ b) How can we define \(d_{0}\) so that the result in part (a) is valid for \(n \geq 2\) ? c) Rewrite the result in part (a) as $$ d_{n}-n d_{n-1}=-\left[d_{n-1}-(n-1) d_{n-2}\right] $$ How can \(d_{n}-n d_{n-1}\) be expressed in terms of \(d_{n-2}, d_{n-3}\) ? d) Show that \(d_{n}-n d_{n-1}=(-1)^{n}\). e) Let \(f(x)=\sum_{n=0}^{\infty}\left(d_{n} x^{n}\right) / n !\). After multiplying both sides of the equation in part (d) by \(x^{n} / n !\) and summingfor \(n \geq 2\), verify that \(f(x)=\left(e^{-x}\right) /(1-x)\). Hence \(d_{n}=n !\left[1-\frac{1}{1 !}+\frac{1}{2 !}-\frac{1}{3 !}+\cdots+\frac{(-1)^{n}}{n !}\right]\)
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