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Chapter 3: Problem 15

Establish the equality \\[ \sum_{\text {diagonal}=1}^{n-1}[(n-\text {diagonal}) \times \text {diagonal}]=\frac{n(n-1)(n+1)}{6} \\] This is used in the every-case time complexity analysis of Algorithm 3.6.

Short Answer

Expert verified
The equality \(\sum_{\text {diagonal}=1}^{n-1}[(n-\text {diagonal}) \times \text {diagonal}]=\frac{n(n-1)(n+1)}{6}\) holds as both sides of the equation represent the same mathematical concept: the sum of products of consecutive numbers for a given range.

Step by step solution

01

Expand the summation term

First, let's expand the terms in the summation for a few terms to understand the pattern. When \(\text{diagonal}=1\), the term will be \((n-1)\times1\), when \(\text{diagonal}=2\), the term will be \((n-2)\times2\), and so on until \(\text{diagonal}=n-1\), where the term will be \((n-(n-1))\times(n-1)=1\times(n-1)\). This forms a pattern of decreasing singlets from \(n-1\) to \(1\), multiplied by increasing singlets from \(1\) to \(n-1\).
02

Use the formula for the sum of consecutive integers

The sum of consecutive integers from \(1\) to \(n-1\) can be given by \(\frac{(n-1)(n-1+1)}{2}\) or \(\frac{n(n-1)}{2}\). The square of this result gives the sum of every pair of products of these singlets.
03

Express the summation as the square of sum of consecutive integers

If we express the summation in terms of sum of consecutive integers, it is the square of the sum, which is \(\left(\frac{n(n-1)}{2}\right)^2\).
04

Compare with the right-hand side

The right-hand side of the given expression is \(\frac{n(n-1)(n+1)}{6}\). This expression represents a well-known formula for the sum of squares of first n-1 natural numbers.
05

Establish equality

The equality can be established by realizing that \(\left(\frac{n(n-1)}{2}\right)^2\) is equivalent to \(\frac{n^2*(n-1)^2}{2^2}\), which is equivalent to \(\frac{n(n-1)(n+1)}{6}\).

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