Problem 69 - A Graphical Approach to College Algebra

Question

Find the sum for each series. $$\sum_{i=1}^{60}\left(i^{3}-2 i^{2}\right)$$

Step-by-Step Solution

Answer

The sum is 3,199,560.

Step 1: Expand the Series

Write out the expression for the series, expanding the sum to encompass both terms:\[\sum_{i=1}^{60} (i^3 - 2i^2) = \sum_{i=1}^{60} i^3 - \sum_{i=1}^{60} 2i^2\]This separates the sum into two distinct parts that can be handled individually.

Step 2: Calculate the Sum of Cubes

Use the formula for the sum of cubes: \( \sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2 \). For \( n = 60 \), the sum becomes:\[\left( \frac{60 \times 61}{2} \right)^2 = (1830)^2 = 3{,}348{,}900\]

Step 3: Calculate the Sum of Squares

Use the formula for the sum of squares: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \). For \( n = 60 \), this yields:\[\frac{60 \times 61 \times 121}{6} = 74{,}670\]Since the second sum is \( 2 \sum_{i=1}^{60} i^2 \), we multiply by 2:\[2 \times 74{,}670 = 149{,}340\]

Step 4: Combine the Results

Subtract the result of the sum of squares from the sum of cubes:\[3{,}348{,}900 - 149{,}340 = 3{,}199{,}560\]This gives the total sum for the series.

Key Concepts

Sum of Cubes Formula

The sum of cubes is a fascinating concept that can simplify calculations for a series where each term is a cube. To calculate the sum of cubes from 1 to a number \( n \), you use the formula:\[\sum_{i=1}^{n} i^3 = \left( \frac{n(n+1)}{2} \right)^2\]This formula is derived from the fact that the cube of each number can be expressed as the square of the sum of the first \( n \) natural numbers. For instance, to find the sum of cubes up to 60, simply plug in 60 into the formula, as shown in the original problem:\[\left( \frac{60 \times 61}{2} \right)^2 = (1830)^2 = 3,348,900\]This provides a quick and efficient way to determine the sum of a cube series without having to manually cube and add each number. The result indicates the total contribution of cubed terms in the given series.

Sum of Squares Formula

When dealing with squares of numbers, the sum can be calculated using a distinct formula that efficiently simplifies these calculations:\[\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\]This formula is derived from the properties of squares which follow a quadratic pattern. It allows us to find the total sum for squares quickly. For the task at hand in the exercise, with \( n = 60 \), the calculation becomes straightforward:\[\frac{60 \times 61 \times 121}{6} = 74,670\]In the given expression, 2\( i^2 \), it is necessary to multiply the result of the sum of squares by 2, because each square is doubled in the overall sum:\[2 \times 74,670 = 149,340\]This technique allows us to manage large sequences efficiently and get the sum without iterative calculations.

Series Expansion

Series expansion is a method used in mathematics to break down complex expressions into simpler parts. This technique is particularly useful for summing sequences where terms follow a recognizable pattern. In the example provided, the series sum is expressed initially as:\[\sum_{i=1}^{60} (i^3 - 2i^2)\]By expanding the series into its individual components: \( i^3 \) and \(- 2i^2 \), it allows us to apply known formulas separately to each part without losing accuracy.

• **First**, calculate the sum of cubes for the series \( i^3 \).
• **Second**, calculate the sum for \(- 2i^2\) by finding the sum of squares and adjusting accordingly.

This separation simplifies computation by handling smaller, more manageable parts. Once each is computed, the total sum is simply the difference between these two calculated sums:\[3,348,900 - 149,340 = 3,199,560\]Series expansion is a powerful analytical tool in algebra and calculus, allowing for streamlined solutions to complex problems.