Problem 32
Question
In Exercises \(31-34,\) use the summation formulas to rewrite the expression without the summation notation. Use the result to find the sum for \(n=10,100,1000,\) and \(10,000 .\) $$ \sum_{j=1}^{n} \frac{4 j+1}{n^{2}} $$
Step-by-Step Solution
Verified Answer
For \(n=10\), the sum is 22.1, for \(n=100\) the sum is 202.01, for \(n=1000\) the sum is 2002.002, and for \(n=10000\) the sum is 20002.0002.
1Step 1: Split the Summation
Break the summation into two summations, one for \(4j\) and another for \(1\). Due to the linearity of the summation operator, this can be written as \( \sum_{j=1}^{n} 4j/n^2 + \sum_{j=1}^{n} 1/n^2 \).
2Step 2: Use the Summation Formulas
Now, rewrite each summation using summation formulas. \(\frac{4}{n^2} \sum_{j=1}^{n} j + \frac{1}{n^2} \sum_{j=1}^{n} 1 \) which turns into \(\frac{4}{n^2} * \frac{n(n+1)}{2} + \frac{1}{n^2} * n\). Simplify this further to get: \(2(n+1) + 1/n = 2n + 2 + 1/n\).
3Step 3: Compute the Sum for Different Values of n
Insert different values of n into the simplified formula to compute the sum. For \(n=10\), the sum is \(2*10 + 2 + 1/10 = 22.1\), for \(n=100\), the sum is \(2*100 + 2 + 1/100 = 202.01\), for \(n=1000\), the sum is \(2*1000 + 2 + 1/1000 = 2002.002\) and for \(n=10000\), the sum is \(2*10000 + 2 + 1/10000 = 20002.0002\).
Key Concepts
Linearity of SummationSimplification of ExpressionsFinite Sums
Linearity of Summation
When working with summation notation, one of the most useful properties is linearity. This principle allows us to break apart complex summation expressions into simpler, more manageable pieces.
For instance, in the given expression \( \sum_{j=1}^{n} \frac{4j+1}{n^{2}} \), we apply linearity by splitting the sum into two separate terms: \( \sum_{j=1}^{n} \frac{4j}{n^{2}} \) and \( \sum_{j=1}^{n} \frac{1}{n^{2}} \). Using linearity, these can be written as the sum of individual constants and variable parts, making calculation straightforward.
For instance, in the given expression \( \sum_{j=1}^{n} \frac{4j+1}{n^{2}} \), we apply linearity by splitting the sum into two separate terms: \( \sum_{j=1}^{n} \frac{4j}{n^{2}} \) and \( \sum_{j=1}^{n} \frac{1}{n^{2}} \). Using linearity, these can be written as the sum of individual constants and variable parts, making calculation straightforward.
- Linearity means you can distribute (or "factor out") constants across the summation sign.
- It simplifies algebraic manipulation by allowing terms to be separated and computed independently.
- This property helps to manage sums in a systematic way, especially when dealing with series and sequences.
Simplification of Expressions
Simplifying expressions is a crucial step in solving any mathematical problem, especially those involving summations. In our exercise, simplification helps turn an otherwise daunting task into a more straightforward calculation.
After applying the linearity of summation, each part of the expression is simplified separately. For example, the term \( \frac{4}{n^2} \sum_{j=1}^{n} j \) simplifies to \( \frac{4}{n^2} \times \frac{n(n+1)}{2} \), which can be further broken down to \( 2(n+1) \). Similarly, \( \frac{1}{n^2} \sum_{j=1}^{n} 1 \) simplifies to \( \frac{n}{n^2} = \frac{1}{n} \).
After applying the linearity of summation, each part of the expression is simplified separately. For example, the term \( \frac{4}{n^2} \sum_{j=1}^{n} j \) simplifies to \( \frac{4}{n^2} \times \frac{n(n+1)}{2} \), which can be further broken down to \( 2(n+1) \). Similarly, \( \frac{1}{n^2} \sum_{j=1}^{n} 1 \) simplifies to \( \frac{n}{n^2} = \frac{1}{n} \).
- Use known summation formulas to simplify each part individually. For instance, \( \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \).
- Simplifying involves calculating constants first, as they often shape the final outcome.
- Check every step to ensure each part is clearly broken down to its most basic form.
Finite Sums
Finite sums involve adding a fixed number of terms. In our problem, we compute a sum depending on the value of \( n \), which defines how many terms we are adding together.
Once the expression is simplified, finding the finite sum is easy and involves just basic arithmetic. Specifically, the exercise involves replacing \( n \) with specific values such as 10, 100, 1000, and 10,000, and then calculating each outcome.
Once the expression is simplified, finding the finite sum is easy and involves just basic arithmetic. Specifically, the exercise involves replacing \( n \) with specific values such as 10, 100, 1000, and 10,000, and then calculating each outcome.
- Finite sums are helpful to provide exact values for realistic scenarios.
- They are bounded, meaning you know how many terms you'll be handling, which helps in calculation.
- Evaluation involves mere substitution and arithmetic operations, making computation direct.
Other exercises in this chapter
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