Problem 24 - 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 $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=1}^{5}\left(4-k^{2}\right) $$

Step-by-Step Solution

Answer

The sum is -35.

Step 1: Split the Sum

The given sum is \( \sum_{k=1}^{5}(4-k^2) \). We can split this up into two separate sums: \( \sum_{k=1}^{5} 4 - \sum_{k=1}^{5} k^2 \).

Step 2: Evaluate the Constant Sum

The first part of the split sum is \( \sum_{k=1}^{5} 4 \). Since 4 is constant for each value of \( k \), this is simply \( 4 \times 5 = 20 \).

Step 3: Apply the Sum of Squares Formula

The second part of the split sum is \( \sum_{k=1}^{5} k^2 \). Using the formula \( \sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6} \), where \( n=5 \), we calculate \( \frac{5(5+1)(2(5)+1)}{6} = \frac{5 \times 6 \times 11}{6} = 55 \).

Step 4: Subtract the Sum of Squares from the Constant Sum

Now subtract the result of the sum of squares from the constant sum: \( 20 - 55 = -35 \).

Key Concepts

Sum of Natural Numbers

Understanding the sum of natural numbers is essential in mathematics. It’s a straightforward concept but very powerful in calculations.Natural numbers are the numbers we usually count with: 1, 2, 3, and so on. If you wanted to find the total of the first few natural numbers, say from 1 to 5, you would add them up:

• 1 + 2 + 3 + 4 + 5 = 15

Interestingly, there’s a formula that allows you to find this sum without adding each number individually. The formula is:\[\sum_{k=1}^{n} k = \frac{n(n + 1)}{2}\]This formula states that if you want to add up the numbers from 1 to \( n \), you multiply \( n \) by \( n + 1 \), then divide the result by 2. This method is especially useful for adding a long sequence of natural numbers quickly.

Sum of Squares Formula

The sum of squares is another fundamental concept, especially in mathematical studies involving variances or physics.When we talk about squares in mathematics, we mean squaring a number, like this: 1 squared is 1, 2 squared is 4, 3 squared is 9, and so on. The sum of squares is the sum of these square numbers up to a certain limit. For example, to find the sum of squares of the first 5 natural numbers, we use:\[1^2 + 2^2 + 3^2 + 4^2 + 5^2\]To avoid calculating each square separately and then adding, the sum of squares formula helps:\[\sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6}\]This formula requires substituting \( n \) with the highest number in the sequence. For example, with \( n = 5 \), the sum becomes 55. This formula simplifies calculations considerably, especially for longer numerical sequences.

Constant Sum Evaluation

A constant sum is remarkably simple yet crucial in many mathematical applications. This concept involves adding a constant value multiple times in a series.For instance, if you have a constant number 4, and you are asked to evaluate the sum of 4 for 5 terms, you simply multiply:

• 4 \( \times \) 5 = 20

The idea is that when you have a constant number repeated over a series, its sum is just the constant multiplied by the number of terms.This concept can be particularly handy when splitting sums, where each sum part involves a constant. By treating constants separately from variable components, you simplify your algebraic expressions, making calculations more manageable.