Problem 34 - Essential Calculus

Question

In Exercises \(31-34,\) use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sum for \(n=10,100,1000,\) and \(10,000 .\) $$ \sum_{i=1}^{n} \frac{4 i^{2}(i-1)}{n^{4}} $$

Step-by-Step Solution

Answer

The sum of the series for \(n = 10\) is \(0.165\), for \(n=100\) it is \(0.0199\), for \(n=1000\) it's \(0.002\) and for \(n=10,000\) the sum is \(0.0002\).

Step 1: Expression Analysis

Given expression is \(\sum_{i=1}^{n} \frac{4 i^{2}(i-1)}{n^{4}} \). It can be broken down as 4 times the summation of \(\frac{i^{2}(i-1)}{n^{4}} \), over \(i = 1\) to \(n\). Now we are to simplify this summation using summation formulas.

Step 2: Use of Summation Formulas

From the formulas, we know:\[\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\]\[\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\]\[\sum_{i=1}^{n} i^3 = \left[\frac{n(n+1)}{2}\right]^2\]Multiplying these, we get the sum \(\frac{i^{2}(i-1)}{n^{4}} = \frac{n(n+1)(2n+1)}{6}\times\frac{(n-1)n(n+1)}{2n^{4}}\). This simplifies to \(\frac{(n-1)(2n+1)}{6n^{3}}\).

Step 3: Substitute Values of n

Now, substitute the values of n (10, 100, 1000, and 10000) into our simplified equation \(\frac{(n-1)(2n+1)}{6n^{3}}\) one at a time, and calculate the corresponding sums.

Key Concepts

Summation Techniques

Summation techniques involve different methods to simplify and calculate the sum of a series of numbers or expressions. Here, summation notation is initially presented as a compact way to write the sum of a sequence. However, it can be a bit tricky at first glance. This exercise focuses on converting the given summation into a form that is much easier to manage without the notation. This is achieved by identifying patterns and using known summation formulas, which are key to tackling such problems easily.

• The first step is to recognize the sum as a collection of terms that can typically be broken down or simplified individually.
• Using known formulas such as the summation of natural numbers, squares, and cubes can greatly reduce the complexity of the problem.
• For example, in this particular question, the formula for the summation of squares, \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\), plays a critical role in simplifying the expression.

Understanding summation techniques not only helps in easing up the calculation process but also helps you recognize the deeper patterns in mathematical expressions.

Mathematical Expressions

Mathematical expressions are a combination of numbers, variables, and operators that represent a quantity or a concept. The main goal when dealing with mathematical expressions in calculus is often to simplify them so that they are easy to evaluate. In the given exercise, we start with the expression \(\sum_{i=1}^{n} \frac{4 i^{2}(i-1)}{n^{4}}\). To handle such expressions, follow these steps:

• Break down complex expressions into smaller parts. Here, identify the component parts as \(i^2\times(i-1)\) to simplify handling it individually.
• Simplify the fraction where possible. Deconstruct \(\frac{4 i^{2}(i-1)}{n^{4}}\) by separating constant factors and incorporating into a summation.
• Use algebraic identities and properties of numbers, which provide tools like the expansion of \(i^2\times(i-1)\), essential in reducing terms.

Algebraic manipulation is a powerful tool in mathematics, streamlining the process and making complex ideas approachable.

Calculus Problem Solving

Calculus problem solving often involves breaking down expressions with sums and limits into more manageable parts, as seen in this exercise. When solving calculus problems, especially those with a series or sequence, it becomes crucial to handle large values with finesse and accuracy.Let's see how:

• First, simplify the summation using known formulas, like the squares and cubes formulas highlighted earlier.
• Substitute known values of \(n\) such as 10, 100, 1000, and 10,000 after simplifying the general expression \(\frac{(n-1)(2n+1)}{6n^{3}}\).
• Calculate the result for these values directly, ensuring each step is straightforward and verifiable. This methodical substitution ensures the final results are both accurate and clear.

The power of calculus lies in its ability to deal with complicated expressions by applying limits, making it possible to solve and understand them in real-world applications. Employing these techniques is vital for higher-level problem solving.