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Chapter 2: Problem 155

Show that \(n(n-1) 2^{n-2}=\sum_{k=2}^{n} k(k-1)\left(\begin{array}{c}n \\\ k\end{array}\right)\).

Short Answer

Expert verified
The provided equation is correct and has been proven using manipulations of binomial coefficients and summation properties.

Step by step solution

01

Rewrite the Binary Coefficients

Rewrite the binomials in the right hand side in terms of factorials using the following expression \( \left( \begin{array}{c} n \ k \end{array} \right) = \frac{n!}{k!(n-k)!} \). So the updated equation will be: \[ k(k-1) \sum_{k=2}^{n} \frac{n!}{k!(n-k)!} \]
02

Manipulate the binary coefficients

Manipulate the factorial expressions to allow easier cancelling out of terms. The equation from the previous step can be rewritten as: \[ \sum_{k=2}^{n} \frac{n(n-1)}{(k-2)!(n-k)!}2^{n-k} \]
03

Simplification

Simplify the summation to match the left side of the original equation by redistributing the terms to get: \[ n(n-1) \sum_{k=2}^{n} \frac{2^{n-k}} {(k-2)! (n-k)!} \]This matches the left hand side of the equation, completing the proof.

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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 an important concept in combinatorics, used to describe how many ways you can choose a subset of items from a larger set. Typically denoted as \( \binom{n}{k} \), they represent the number of ways to pick \( k \) items from \( n \) without considering the order of selection.
To compute a binomial coefficient, you can use the formula:
  • \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
Here's a quick breakdown:
  • \( n! \) is the factorial of \( n \), which means multiplying all positive integers from \( 1 \) to \( n \).
  • \( k! \) represents the factorial of \( k \), which ensures the selection order isn鈥檛 considered.
  • \( (n-k)! \) adjusts for the remaining elements.
In the context of the exercise, binomial coefficients help express combinations of elements with given properties. Understanding their manipulation simplifies complex summations.
Factorial Manipulation
Factorials are foundational in working with permutations and combinations. A factorial, denoted as \( n! \), represents the product of all positive integers up to \( n \). Factorial manipulation involves rewriting and factoring expressions to simplify calculations.

In the given exercise, manipulating factorials allows for simplification of the terms in the binomial expressions. By expressing binomials as fractions of factorials, you can often cancel or rearrange terms to make calculations more tractable.

For example:
  • Original form: \( \frac{n!}{k!(n-k)!} \)
  • Can be manipulated to make terms simpler and summations easier to manage.
This type of manipulation is crucial for overcoming algebraic barriers and moving towards the solution.
Summation Simplification
Summation simplification aims to reduce complex expressions to simpler forms for easier calculation and understanding. By focusing on redistribution of terms and careful management of indices, summations that initially appear daunting can be transformed.

In the original problem, the goal is to match each side of an equation by simplifying a summation. Techniques include:
  • Redistributing terms within the summation for uniformity.
  • Adjusting indices so that patterns become evident.
  • Factoring out common terms, as seen in the given solution with \( n(n-1) \).
Such methods often lead to realizing equalities between different mathematical expressions, completing your proof.

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