Problem 10 - Precalculus with Limits

Question

In Exercises 5-12, evaluate the sum using the summation formulas and properties. $$\displaystyle\sum_{k=1}^{50} (2k + 1)$$

Step-by-Step Solution

Answer

Therefore, the sum of the series \(\sum_{k=1}^{50} (2k + 1)\) is \(2550 + 50 = 2600\).

Step 1: Split the sum into two sums

Use the properties of summations to split the original sum into two separate sums: \(\sum_{k=1}^{50} (2k + 1) = \sum_{k=1}^{50} 2k + \sum_{k=1}^{50} 1\)

Step 2: Pull constants out of the sum

Use the properties of summations again to pull any constants out of the sums. The 2 can be taken out of the first sum, giving: \(\sum_{k=1}^{50} 2k = 2 \sum_{k=1}^{50} k\) and the second sum remains the same as \( \sum_{k=1}^{50} 1 \)

Step 3: Apply the sum formulas

Now, you can apply the sum formulas. For the first sum, use the formula for the sum of the first \(n\) integers, which is \(\frac{n(n + 1)}{2}\), so the sum \(2 \sum_{k=1}^{50} k = 2 \cdot \frac{50 \cdot (50 + 1)}{2}\). For the second sum, since you are adding up 1 fifty times, this simplifies to 50.

Step 4: Evaluate the sums

Finally, you can evaluate the two sums. For the first sum, this results in \(2 \cdot \frac{50 \cdot (50 + 1)}{2} = 2550\). The second sum is just 50. Therefore, the sum of these two results gives the total sum.

Key Concepts

Summation Properties

Summation properties make working with sums much easier by allowing you to manipulate and simplify expressions. These properties include linearity, where you can split a sum into multiple simpler parts.
For example, if you have a sum of two terms like \( \sum_{k=1}^{n} (a_k + b_k) \), you can split it into \( \sum_{k=1}^{n} a_k + \sum_{k=1}^{n} b_k \).
This is particularly useful in our original exercise, where the term \( 2k + 1 \) was split into separate sums.

• This separation makes it easier to deal with each part using known formulas.
• Another useful property is factoring constants out of sums, which simplifies calculations.

For instance, \( \sum_{k=1}^{n} c \cdot a_k \) becomes \( c \cdot \sum_{k=1}^{n} a_k \). These properties are fundamental because they transform complex expressions into manageable ones, preparing them for using summation formulas.

Sum of Integers

The sum of the first \( n \) integers is a well-known formula: \( \frac{n(n+1)}{2} \).
This formula is easy to apply and powerful for solving various mathematical problems involving sums.
In the example given, to calculate \( \sum_{k=1}^{50} k \), we simply substitute \( n = 50 \) into our formula, resulting in \( \frac{50(50 + 1)}{2} \).

• The computation is straightforward: 50 multiplied by 51 equals 2550, which when divided by 2 gives 1275.

This approach provides a quick and accurate means to find the sum of consecutive integers, avoiding lengthy additions.

Evaluation of Sums

Evaluating sums often involves applying specific formulas or breaking the sum into simpler components.
The initial step generally requires applying summation properties to simplify a complex sum into manageable parts.
In our case, the sum \( \sum_{k=1}^{50} (2k + 1) \) is split and then constants are factored out.

• The first sum \( 2 \sum_{k=1}^{50} k \) uses the formula for the sum of integers, while the second sum \( \sum_{k=1}^{50} 1 \) evaluates to 50, since it's just adding "1" fifty times.
• By calculating these sums separately, you can easily evaluate the entire expression.

This methodical approach simplifies complex operations by addressing them in smaller, simpler steps.

Mathematical Series

A mathematical series is essentially the sum of terms in a sequence.
When dealing with a finite series, like the one presented in the exercise, we use a combination of summation properties and specific formulas to find the total sum.
In mathematical series, each term in the sum can often be expressed in a general form, where recognizing patterns helps in applying the right formulas.

• For instance, the series \( 2k + 1 \) in our exercise is a linear arithmetic pattern, which we can handle using known techniques and properties.

Understanding the nature and structure of a series aids significantly in its evaluation. Recognizing patterns allows one to apply the correct summation formula, greatly speeding up calculations and ensuring accuracy.