Problem 70 - A Graphical Approach to College Algebra

Question

Find the sum for each series. $$\sum_{i=1}^{43}\left(15 i^{2}-2\right)$$

Step-by-Step Solution

Answer

The sum is 411424.

Step 1: Identify the Series Expression

The series given is \( \sum_{i=1}^{43} (15i^2 - 2) \). It involves two parts: \( 15i^2 \) and \(-2 \). Both parts will be summed separately.

Step 2: Separate Components of the Sum

Break the series into two separate sums: \( \sum_{i=1}^{43} 15i^2 \) and \( \sum_{i=1}^{43} (-2) \). This allows us to handle each part individually.

Step 3: Sum the Quadratic Component

Use the formula for the sum of squares: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \). For \( n = 43 \), substitute to find: \( \sum_{i=1}^{43} i^2 = \frac{43 \cdot 44 \cdot 87}{6} = 27434 \). Then multiply by 15: \( 15 \cdot 27434 = 411510 \).

Step 4: Sum the Constant Component

For the constant \( -2 \), the sum is simply \( -2 \) multiplied by the number of terms. With 43 terms, the sum is \( -2 \cdot 43 = -86 \).

Step 5: Combine the Results

Add the results from Steps 3 and 4 together: \( 411510 + (-86) \). The final result is \( 411424 \).

Key Concepts

Quadratic Series

A quadratic series involves terms that are quadratic in nature. In simpler terms, each term in the series can be represented in the form of a quadratic equation, such as \( ax^2 + bx + c \). In this exercise, the quadratic component is \( 15i^2 \). The sum of a quadratic series can be more challenging to compute than an arithmetic series due to the presence of the \( i^2 \) term.
By understanding how these series work, we can tackle problems that include quadratic terms. Such series are often split to make calculations easier, as we did in the exercise. Separating the quadratic term allows us to focus on it exclusively, using specialized formulas to simplify our work.

Sum of Squares Formula

The sum of squares formula is pivotal when working with quadratic series. It allows us to calculate the sum of squared terms like \( i^2 \) from 1 to \( n \). The formula is \[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \].
This formula is derived from summing the squares of natural numbers. In our exercise, applying this formula to calculate \( \sum_{i=1}^{43} i^2 \) involves substituting 43 for \( n \), resulting in a sum of 27,434.
After calculating the sum of squares, the next step involves multiplying it by the coefficient of the squared term, in this case, 15. This multiplication gives us 411,510, which represents the sum of the quadratic series component. Utilizing this formula helps handle complex series and convert them into simpler, calculable expressions.

Constant Term in Series

The constant term in a series refers to a term that remains the same in every sequence of the series. Here, the constant term is \(-2\).
To find the sum of the constant term in the series, we simply multiply the constant by the number of terms. For example, with \(-2\) repeated over 43 terms, the sum calculation becomes \(-2 \times 43 = -86\).
This straightforward approach saves time and effort, especially when combined with more intricate components of a series. By adding this calculated constant term sum to the result of other components, we achieve the complete series sum, totaling 411,424 in the exercise. Understanding constant terms helps ensure all components are accounted for when solving series problems.