Problem 6 - Calculus Volume 1

Question

Suppose that \(\sum_{i=1}^{100} a_{i}=15\) and \(\sum_{i=1}^{100} b_{i}=-12 .\) In the following exercises, compute the sums. $$\sum_{i=1}^{100}\left(3 a_{i}-4 b_{i}\right)$$

Step-by-Step Solution

Answer

The sum is 93.

Step 1: Understand the problem statement

We are given two separate sums: \( \sum_{i=1}^{100} a_{i} = 15 \) and \( \sum_{i=1}^{100} b_{i} = -12 \). We need to find the value of \( \sum_{i=1}^{100} (3a_i - 4b_i) \).

Step 2: Use properties of summation to separate terms

The expression \( \sum_{i=1}^{100} (3a_i - 4b_i) \) can be separated using summation properties into two separate sums: \( 3\sum_{i=1}^{100} a_i - 4\sum_{i=1}^{100} b_i \).

Step 3: Substitute known values into the new expression

Substitute the known values of \( \sum_{i=1}^{100} a_i \) and \( \sum_{i=1}^{100} b_i \) into the expression. This gives us \( 3(15) - 4(-12) \).

Step 4: Simplify the expression

Calculate each term: \( 3 \times 15 = 45 \) and \( -4 \times (-12) = 48 \). Now sum these values: \( 45 + 48 \).

Step 5: Calculate the final sum

Add the results from Step 4: \( 45 + 48 = 93 \). This is the value of \( \sum_{i=1}^{100} (3a_i - 4b_i) \).

Key Concepts

Summation Properties

Understanding the properties of summation is crucial when dealing with complex sums and series.These properties allow us to simplify expressions and compute sums more easily, especially when multiple terms are involved.

When we see an expression like \( \sum_{i=1}^{100} (3a_i - 4b_i) \), one of the key properties we use is **linearity of summation**.This means:

• We can separate the summation of a linear combination of terms into the sum of each term separately. For example, \( \sum (3a_i - 4b_i) \) can be split into \( 3\sum a_i \) and \( -4\sum b_i \).
• We can factor constants out of sums. This allows us to simplify 3\sum a_i into 3 \times \sum a_i.

These properties make it much simpler to handle and simplify complex expressions by breaking them into smaller, more manageable parts.

Simplifying Expressions

Simplifying expressions is an essential skill in calculus, particularly when we have sums involved.In our exercise, the expression \( \sum_{i=1}^{100} (3a_i - 4b_i) \) is simplified using known sums.

Steps to simplify include:

• Using the **separation property** to break the sum into \( 3\sum a_i - 4\sum b_i \).
• Substituting known values, where \( \sum a_i = 15 \) and \( \sum b_i = -12 \), into the equation.
• Computing each individual part, which are \( 3(15) \) and \( -4(-12) \).

This process makes the problem more straightforward by gradually reducing a complex expression into simple arithmetic operations.

Arithmetic Operations

Arithmetic operations are the foundation of simplifying and solving the sums.Here, operations include multiplication and addition, which are directly applied after using summation properties.

In our solution:

• We multiplied constants and the result of sums, for example, calculating \( 3 \times 15 = 45 \).
• Multiplying negatives, \(-4 \times (-12) = 48\), helps clarify how negative values affect the sign of the result.
• Finally, we sum these results, \( 45 + 48 \), to get the final result of 93.

Mastering these operations is crucial in calculus, as they form the basis for handling much larger and more complex expressions.