Problem 33 - 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 \(10,000 .\) $$ \sum_{k=1}^{n} \frac{6 k(k-1)}{n^{3}} $$

Step-by-Step Solution

Answer

Calculate each sum individually for different 'n' values and then simplify. The process requires understanding of summation and basic algebra.

Step 1: Understand the Formula

We have a summation formula \( \sum_{k=1}^{n} \frac{6 k(k-1)}{n^{3}} \). This equation implies summing up the function \(f(k) = \frac{6 k(k-1)}{n^{3}}\) for k values ranging from 1 to n.

Step 2: Eliminate the Summation Notation

Let's find the total sum when \(n = 10, 100, 1000, 10000\). For \(n = 10\), the sum S would be the sum of \(f(k)\) for all k values from 1 to 10, computed individually and added together.

Step 3: Calculate for n = 10

Substitute k = 1 to 10 in the function \(f(k) = \frac{6 k(k-1)}{10^{3}}\) and add all the obtained values. Repeat this step for other values of n (100, 1000, 10000).

Step 4: Simplify the Results

After computing the sums for all n values, simplify the results to their lowest form.

Key Concepts

Series and Sequences

A sequence is an ordered list of numbers where each number is called a term. When you add up the terms of a sequence, it becomes a series. Understanding sequences and series is crucial for mastering summation formulas. The exercise involves examining a series where each term is represented by a function \( f(k) = \frac{6k(k-1)}{n^3} \), and the goal is to compute the sum of these terms as \( k \) ranges from 1 to \( n \).

Summations allow for the compact representation of adding many terms, using the sigma notation \( \sum \). In this exercise, you're required to rewrite the given expression without this notation and evaluate for multiple values of \( n \). Understanding how to work with sequences and series is foundational for calculus and other advanced mathematical topics.

Calculus Problems

Calculus often involves finding solutions to problems using techniques like differentiation and integration. Summation problems like the one here bridge the gap between algebra and calculus, giving a deeper understanding of how to deal with functions and their sums.

With \( \sum_{k=1}^{n} \frac{6k(k-1)}{n^3} \), you're dealing with a discrete function. Calculus problems sometimes transform into dealing with continuous counterparts through limits, but here, you focus on finite sums. This approach helps in understanding convergence and divergence of series when \( n \) approaches infinity or other large numbers. Tackling different \( n \) values helps solidify these concepts.

Step-by-Step Solutions

Solving problems step-by-step can make seemingly complex calculations more manageable. Here's how you can approach this problem:

• Understand the Formula: Recognize what each part of \( \frac{6k(k-1)}{n^3} \) signifies. Here, \( 6k(k-1) \) represents the mathematical manipulation inside the sequence, and \( n^3 \) stays constant for summing across \( k \).
• Eliminate Summation Notation: Break down the summation into a straightforward calculation by substituting each integer from 1 to \( n \) into \( f(k) \).
• Perform Calculations for Different \( n \): Conduct the process for \( n = 10, 100, 1000, 10000 \), ensuring precision in each step to prevent errors.

By solving sequentially, you achieve a clearer understanding and avoid the pitfalls of trying to calculate everything at once. Each step builds on the previous one, leading to a final refined answer.