Problem 23 - Calculus for Biology and Medicine

Question

Use the algebraic rules for sums to evaluate each sum. Recall that $$\sum_{k=1}^{n} k=\frac{n(n+1)}{2}$$ and $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=1}^{15}(2 k+3) $$

Step-by-Step Solution

Answer

The sum is 285.

Step 1: Expand the Sum

To solve the given sum \( \sum_{k=1}^{15}(2k+3) \), we start by expanding it into two separate sums, corresponding to the linear and constant terms: \( \sum_{k=1}^{15} 2k + \sum_{k=1}^{15} 3 \).

Step 2: Factor Out Constants

In each expanded sum, factor out the constants: \( 2 \sum_{k=1}^{15} k \) and \( 3 \cdot \sum_{k=1}^{15} 1 \).

Step 3: Evaluate the Linear Sum

Use the formula for the sum of the first \( n \) natural numbers: \( \sum_{k=1}^{15} k = \frac{15(15+1)}{2} = 120 \). Then multiply by 2: \( 2 \times 120 = 240 \).

Step 4: Evaluate the Constant Sum

The constant sum is the total of adding 3 a total of 15 times, making it \( 3 \times 15 = 45 \).

Step 5: Combine Both Results

Add the results of the linear and constant sums: \( 240 + 45 = 285 \).

Key Concepts

Sum of Natural Numbers

Natural numbers are simply the numbers that we generally use to count, i.e., 1, 2, 3, and so on. The sum of the first \( n \) natural numbers is a known formula: \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \). This formula allows us to quickly calculate the sum without adding each number separately, which is especially useful when dealing with large values of \( n \).

To understand how this works, consider how the formula is derived. By pairing numbers from opposite ends, such as 1 and \( n \), 2 and \( n-1 \), and so on, you find that each pair sums to \( n+1 \). As there are \( n/2 \) such pairs (for even \( n \); for odd \( n \), one number is left unpaired), multiplying gives \( n(n+1)/2 \).

Example:
For \( n = 15 \), the sum is \( \frac{15 \times 16}{2} = 120 \). Instead of manually adding numbers from 1 to 15, this formula provides a quick solution.

Constant Multiplication in Sums

When dealing with sums where each term is multiplied by a constant, you can take the constant outside of the summation sign. This rule simplifies many algebraic expressions.

• Given a sum: \( \sum_{k=1}^{n} C \cdot a_k \), where \( C \) is a constant.
• You can rewrite it as: \( C \sum_{k=1}^{n} a_k \).

In the original exercise, the expression \( 2k \) means "two times \( k \)". You can factor the 2 out of the sum: \( 2 \sum_{k=1}^{15} k \).

This principle makes calculations no longer dependent on manual multiplications and additions for each term. Using our previous sum of the first natural numbers as an example, we find \( 2 \times 120 = 240 \). This drastically reduces the amount of work needed, especially for large sums.

Expansion of Sums

Expanding sums is a fundamental concept in algebra that breaks a complex sum into simpler, easier-to-handle parts. The idea is to separate terms involving different operations. For example, each term in \( 2k + 3 \) can be thought of independently and thus treated separately in the sum.

• The sum \( \sum_{k=1}^{15}(2k+3) \) is broken down to \( \sum_{k=1}^{15} 2k + \sum_{k=1}^{15} 3 \).
• The first sum, \( 2k \), uses the linear component, allowing the use of constant multiplication in sums.
• The second sum, involving 3 alone, is a constant over the sum and scales directly with the number of terms.

This approach, expanding generally non-uniform terms, means managing them according to their individual rules and simplifies the work you need to do. This breakdown makes the effect of each operation easier to compute, and in our example, it was the difference between manually computing each term and utilizing formulas. This sum ends up as \( 240 + 45 = 285 \), combining both expanded results.