Problem 16 - Calculus: Early Transcendental Functions

Question

Use summation rules to compute the sum. $$\sum_{i=1}^{250}\left(i^{2}+8\right)$$

Step-by-Step Solution

Answer

The sum \(\sum_{i=1}^{250}\left(i^{2}+8\right)\) is 20843750.

Step 1: Separating the Sum

Distribute the sum over the addition. This gives: \(\sum_{i=1}^{250}\left(i^{2}+8\right) = \sum_{i=1}^{250}i^{2} + \sum_{i=1}^{250}8\)

Step 2: Computing the Square Sum

To compute the sum of squares, use the formula \(\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}\). Substitute \(n\) for 250 to get: \(\frac{250(250+1)(2*250+1)}{6} = 20841750\)

Step 3: Computing the Constant Sum

The constant sum is simply 250 times 8, which equals 2000.

Step 4: Final Sum

Add the two computed sums: \(20841750 + 2000\) which gives a final result of: \(20843750\)

Key Concepts

Sum of Squares Formula

Understanding the sum of squares formula can greatly simplify the process of adding up the squares of a series of numbers. This particular formula is a powerful tool in algebra and analysis. It expresses the cumulative total of the squares of the first n natural numbers, which otherwise could be a tedious task to calculate individually and then sum up.

The sum of squares formula is traditionally represented as \[ \sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6} \]. When applied, it provides a quick way to find the sum of squares up to any given positive integer n. In our exercise, the formula was applied as \[ \sum_{i=1}^{250}i^{2} = \frac{250(250+1)(2 \cdot 250+1)}{6} = 20841750 \], swiftly handling what would have been a quite substantial calculation.

Constant Series Summation

When dealing with series in mathematics, a constant series refers to a sequence of numbers where each term is the same fixed value. The summation of a constant series over a given number of terms is an elementary yet fundamental concept. It can be easily calculated by multiplying the constant value by the number of terms. This is based on the simple principle that adding a number to itself a certain number of times is equivalent to multiplication.

In our exercise, the sum of a constant series was \[ \sum_{i=1}^{250}8 = 8 \cdot 250 = 2000 \]. This step is crucial because it simplifies computing a series where a constant value is added repeatedly, emphasizing the versatility and efficiency of summation techniques in algebra.

Algebraic Summation Techniques

The realm of algebra is rich with techniques for simplifying and computing series and sequence summations. One of the foundational approaches is the separation of terms, which involves breaking down complex series into simpler components that are easier to manage. This technique employs the distributive property of addition over summation, allowing each part of the series to be handled independently.

In the given problem, we first broke down the original summation of \(i^2 + 8\) into two separate sums. This separation made it manageable to apply the sum of squares formula to the series of squared terms and a basic multiplication for the constant series. Concluding with the addition of these two results yields the final answer to the summation problem. Such algebraic techniques not only facilitate computation but also help in understanding the underlying properties of series and sequences in mathematics.