Problem 35 - Discrete Mathematics with Applications

Question

Evaluate each sum. $$ a\left(\begin{array}{l}{n} \\\ {0}\end{array}\right)+(a+d)\left(\begin{array}{l}{n} \\\ {1}\end{array}\right)+(a+2 d)\left(\begin{array}{l}{n} \\\ {2}\end{array}\right)+\cdots+(a+n d)\left(\begin{array}{l}{n} \\\ {n}\end{array}\right) $$ (Hint: Use the same hint as in Exercise \(34 .\) )

Step-by-Step Solution

Answer

In short, the sum can be evaluated using the properties of arithmetic progression and binomial coefficients as follows: $$ S_n = \sum_{k=0}^n (a+kd)\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) $$ Use the hint from Exercise 34 and introduce a dummy term 'r': $$ S_n = \sum_{k=0}^n [(a+r) + (k-r)d]\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) $$ Separate the sum into two parts: $$ S_n = (a+r)2^n + n(d-r)2^{n-1} $$ This general expression can help evaluate the sum for any given values of a, d, n, and r.

Step 1: Identify the arithmetic progression and binomial coefficient terms

We can see that the terms in the sum involve an arithmetic progression with first term 'a' and common difference 'd', and binomial coefficients, written as $\left(\begin{array}{l}{n} \\\ {k}\end{array}\right)$, where n is constant and k varies from 0 to n.

Step 2: Write the sum in a general form

Let's write the sum in a general form using the arithmetic progression and binomial coefficient terms: $$ S_n = \sum_{k=0}^n (a+kd)\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) $$

Step 3: Introduce a dummy term `r` to simplify the expression

Using the hint from Exercise 34, introduce a dummy term `r` and add and subtract it from the sum's terms, then group the resulting terms that have the binomial coefficient common. $$ S_n = \sum_{k=0}^n [(a+r) + (k-r)d]\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) $$

Step 4: Separate the suminto two sums, each having a binomial coefficient

Now separate the sum into two sums, which can be simplified using the binomial coefficient properties: $$ S_n = \sum_{k=0}^n (a+r)\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) + \sum_{k=0}^n (k-r)d\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) $$

Step 5: Find the sum for each of the two parts using binomial coefficient properties

We know that the sum of the binomial coefficients is \(2^n\) and the sum of the product of the binomial coefficient and the index is n\(2^{n-1}\). Use these properties to find the sums of the two parts: $$ \sum_{k=0}^n (a+r)\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) = (a+r)2^n $$ $$ \sum_{k=0}^n (k-r)d\left(\begin{array}{l}{n} \\\ {k}\end{array}\right) = n(d-r)2^{n-1} $$

Step 6: Add both parts and find the general expression for S_n

Now, add both parts to get the final expression for S_n: $$ S_n = (a+r)2^n + n(d-r)2^{n-1} $$ With this general expression, we can find S_n for any given values of a, d, n and r. Note that if you have more information about the relationship between r and d, you can simplify this expression further for specific cases.

Key Concepts

Arithmetic Progression

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a constant, known as the common difference, to the previous term. This common difference can be positive, negative, or even zero, leading to an increasing, decreasing, or constant sequence, respectively.

For instance, in the sequence 3, 7, 11, 15, ..., the common difference is 4, as each term is 4 more than the previous one. The nth term of an arithmetic progression can be expressed as:
\( a_n = a_1 + (n - 1)d \)
where \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the position of the term in the sequence. Understanding arithmetic progressions is vital for solving complex summations involving sequences and series, as often encountered in algebra and calculus.

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that provides a quick way of expanding expressions that are raised to a power. It is represented by:
\( (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)
where \( \binom{n}{k} \) (read as 'n choose k') are the binomial coefficients, n is a non-negative integer and k varies from 0 to n. These coefficients can be found in Pascal's Triangle or calculated using the formula:
\( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
The Binomial Theorem is extensively used in probability, statistics, and various areas of mathematics and science. In the context of the given exercise, binomial coefficients are part of the terms being summed, indicating the theorem's relevance in expressing the sum of products with varying powers.

Summation

The concept of Summation is a fundamental aspect of mathematics that involves finding the total or combined value of a sequence of numbers, functions, or terms.

A summation is often written using the sigma notation, which provides a compact way to represent the addition of a sequence of terms:
\( S = \sum_{i=m}^{n} a_i \)
where \( S \) is the sum, \( m \) and \( n \) are the limits of summation, and \( a_i \) represents the terms in the sequence. The goal is to find the sum of all terms \( a_i \) where i starts at \( m \) and ends at \( n \). Summation methods can vary greatly depending on the sequence's properties and nature, such as whether it forms an arithmetic progression, geometric progression, or involves binomial coefficients like in our exercise.

Understanding how to manipulate and calculate sums is crucial in various fields of study, including calculus, analysis, and even computer science, where algorithms often require efficient summation techniques.