Problem 95 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the sum. $$\sum_{i=2}^{5}\left[(i-1)^{3}+(i+1)^{2}\right]$$

Step-by-Step Solution

Answer

The sum of the series is 186

Step 1: Write out the individual terms of the series

The first step is to expand the terms of the summation. The terms of the series are calculated by replacing the variable 'i' with each integer from 2 to 5:Term 1: \( (2-1)^{3} + (2+1)^{2} = 1^3 + 3^2 = 1 + 9 = 10 \), Term 2: \( (3-1)^{3} + (3+1)^{2} = 2^3 + 4^2 = 8 + 16 = 24 \), Term 3: \( (4-1)^{3} + (4+1)^{2} = 3^3 + 5^2 = 27 + 25 = 52 \), Term 4: \( (5-1)^{3} + (5+1)^{2} = 4^3 + 6^2 = 64 + 36 = 100 \)

Step 2: Find the sum of the terms

The second step is to simply add together all the terms:Total sum = Term 1 + Term 2 + Term 3 + Term 4 = 10 + 24 + 52 + 100 = 186

Key Concepts

Series Expansion

When dealing with summation problems in algebra, series expansion is a crucial step. It involves writing out each term of the series by substituting the index variable with each of its specified values. Let's say you have a series like \( \sum_{i=2}^{5}\left[(i-1)^{3}+(i+1)^{2}\right] \). This indicates that you need to expand the series for each integer value from 2 to 5. Break it down term by term:

• For \( i = 2 \), compute \( (2-1)^3 + (2+1)^2 \).
• For \( i = 3 \), compute \( (3-1)^3 + (3+1)^2 \).
• For \( i = 4 \), compute \( (4-1)^3 + (4+1)^2 \).
• For \( i = 5 \), compute \( (5-1)^3 + (5+1)^2 \).

By expanding each term individually, you can easily work out the contribution of each to the total sum. This method also ensures that no terms are overlooked. It makes tackling algebraic summation problems more systematic. Such a detailed approach helps verify that you have calculated each term correctly.

Arithmetic Operations

Arithmetic operations are the basic calculations you perform when expanding a series or solving any mathematical problem. They involve addition, subtraction, multiplication, and division. In the context of series expansion, we mainly focus on addition and multiplication, but subtraction and division can come up as well.
To solve the summation problem, treat each expression separately:

• First, handle the subtraction or addition within the parentheses, like \( (i-1) \) or \( (i+1) \).
• Next, apply multiplication operations when you compute powers. For example, for \( (i-1)^3 \), multiply \( (i-1) \) by itself twice.
• Finally, add the results together as described in your mathematical expression.

These operations must be performed in the proper order: parentheses, multiplication and division, addition and subtraction, according to the mathematical order of operations, often abbreviated as PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). Following these steps correctly leads to simplified and accurate solutions.

Exponents

Exponents are a way to express repeated multiplication compactly. They play a significant role in algebra, especially in problems involving series expansion. An exponent tells you to multiply the base number by itself a certain number of times. For instance, in the summation \( (i-1)^3 \), the base is \( i-1 \), and the exponent is 3. This means you calculate \( (i-1) \times (i-1) \times (i-1) \).
Exponents follow specific rules:

• Product of Powers: Multiply exponents with the same base by adding their exponents.
• Power of a Power: Multiply exponents when raising a power to another power, such as \((a^b)^c = a^{b \cdot c}\).
• Negative Exponents: A negative exponent implies reciprocal power, \( a^{-n} = \frac{1}{a^n}\).
• Zero Exponent: Any non-zero number raised to the zero power equals 1, \( a^0 = 1\).

Understanding these rules helps in simplifying expressions involving exponents, ensuring accurate results in your algebra problems. You can also curb mistakes by checking your expansions against these rules.