/*! 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 6 Find a closed form for the summa... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a closed form for the summation \(\sum_{k=0}^{n} k\left(\begin{array}{l}n \\\ k\end{array}\right)\).

Short Answer

Expert verified
The closed form is \( n \cdot 2^{n-1} \).

Step by step solution

01

Identify the Problem Type

Recognize that we need to find a closed-form expression for the summation \( \sum_{k=0}^{n} k \left( \begin{array}{c} n \ k \end{array} \right) \). This problem involves a combination of binomial coefficients and a linear term in \(k\).
02

Use Known Binomial Identity

Recall the binomial identity \( \sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1} \). This identity allows us to express the series in a closed form.
03

Apply the Identity

Directly apply the binomial identity to find the closed form:\[\sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1}\]
04

Verification

Verify the result by considering small values of \(n\), such as \(n=0\), \(n=1\), and \(n=2\). Compute each summation and check if they match \(n \cdot 2^{n-1}\). For each case, the identity should hold true.

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Key Concepts

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

Binomial Coefficients
Binomial coefficients are fundamental in combinatorics. They appear in the expansion of expressions raised to a power, like \[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k \]The binomial coefficient \( \binom{n}{k} \) counts the number of ways to choose \( k \) elements from \( n \) elements without considering the order. It is calculated as:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]where \( n! \) is the factorial of \( n \). The presence of the binomial coefficient in problems like the one in the exercise helps bridge the gap between algebraic expansions and counting principles in discrete mathematics.
Understanding these coefficients is crucial because they simplify complex summations and offer insights into combinatorial problems. Using identities associated with them, like the one stated in the solution \( \sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1} \), can significantly simplify the process of finding closed-form expressions.
Summation Techniques
Summation techniques are powerful tools in mathematics that help calculate the total of a sequence of terms. When dealing with series that involve binomial coefficients, it is helpful to use known identities and simplifications.One such technique involves recognizing patterns or identities that fit the given summation. In our original exercise, leveraging the identity \( \sum_{k=0}^{n} k \binom{n}{k} = n \cdot 2^{n-1} \) was key to simplifying the problem.
This technique saves time and reduces calculation errors, allowing you to apply a directly known result rather than computing every single term of a series manually. By mastering summation techniques, you can develop solutions that are not only quicker but also more elegant, fostering a deeper understanding of mathematical concepts.
Discrete Mathematics
Discrete mathematics deals primarily with finite or discrete elements. It stands apart from continuous mathematics, which involves objects in a continuous flow or spectrum. Discrete mathematics is fundamental in computer science and combinatorics. Topics in discrete mathematics include logic, algorithms, graphs, and especially combinatorics – the study of counting. In exercises like the one provided, elements of discrete mathematics manifest themselves in the form of counting combinations and patterns using binomial coefficients, showing how they are applied in forming closed-form expressions.
The famous "n choose k" calculation is a discrete math tool used in problems ranging from probability to network theory. This highlights the interconnectivity of mathematical disciplines using techniques from discrete mathematics to solve problems across a broad range of areas including cryptography and computer algorithms. Working with discrete math problems enhances analytical skills and problem-solving capabilities.

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Most popular questions from this chapter

A checkerboard has 64 distinct squares arranged into eight rows and eight columns. (a) In how many ways can eight identical checkers be placed on the board so that no two checkers can occupy the same row or the same column? (b) In how many ways can two identical red checkers and two identical black checkers be placed on the board so that no two checkers of the same color can occupy the same row or the same column?

How many five-letter words (technically, we should call them strings, because we do not care if they make sense) can be formed using the letters \(A, B, C,\) and \(D,\) with repetitions allowed. How many of them do not contain the substring BAD?

Bridget has \(n\) friends from her bridge club. She is able to invite a different subset of three of them to her home every Thursday evening for 100 weeks. What is the minimum value of \(n ?\)

How many eight-character passwords can be formed with the 26 letters in the English alphabet, each of which can be in uppercase or lowercase, and the 10 digits? How many of them do not have repeated character?

The objective of this problem is to derive a formula for \(\sum_{k=1}^{n} k^{3}\). (a) Use induction to show that $$ \sum_{k=1}^{n}\left(\begin{array}{l} k \\ 3 \end{array}\right)=\frac{n(n+1)(n-1)(n-2)}{4 !} $$ for any positive integer \(n\). Hint: Note that \(\left(\begin{array}{l}1 \\ 2\end{array}\right)=0\) (b) Find the integers \(a, b,\) and \(c\) such that $$ k^{3}=a\left(\begin{array}{l} k \\ 3 \end{array}\right)+b\left(\begin{array}{l} k \\ 2 \end{array}\right)+c\left(\begin{array}{l} k \\ 1 \end{array}\right) $$ Hint: Compare coefficients. (c) Apply the results from parts (a) and (b) to derive a formula for \(\sum_{k=1}^{n} k^{3}\).

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