Problem 28 - 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 $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=1}^{n}(k+2)(k-2) $$

Step-by-Step Solution

Answer

The sum is \( \frac{n(n+1)(2n+1)}{6} - 4n \).

Step 1: Expand the Expression

The given expression is \( (k+2)(k-2) \). This is a difference of squares, which can be expanded using the identity \( a^2 - b^2 = (a-b)(a+b) \). Therefore, \( (k+2)(k-2) = k^2 - 4 \).

Step 2: Split the Sum Into Two Separate Sums

Now that we have \( k^2 - 4 \), we can split the original sum \( \sum_{k=1}^{n}(k+2)(k-2) \) into two separate sums: \( \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} 4 \).

Step 3: Evaluate Each Part of the Sum

Using the given formulas: \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \) and the constant sum \( \sum_{k=1}^{n} 4 = 4n \). Plugging these into the expression from Step 2 gives us: \( \frac{n(n+1)(2n+1)}{6} - 4n \).

Step 4: Simplifying the Result

Simplify the expression: \( \frac{n(n+1)(2n+1)}{6} - 4n \). Find a common denominator and simplify the expression where necessary to get a clearer final form.

Key Concepts

Difference of Squares

The difference of squares is a useful algebraic identity that simplifies expressions of the form \((a+b)(a-b)\). The identity is expressed as \(a^2 - b^2\). Here, you are essentially subtracting the square of one number from the square of another.
This formula is useful because it transforms products into differences, making calculations simpler. For our exercise, the expression \((k+2)(k-2)\) can be expanded into \(k^2 - 4\), since \(b^2\) in this context is \(2^2 = 4\).
The intuitive part of understanding this is to recognize you're dealing with two symmetric binomials, adding and subtracting the same value from a base \(k\). This symmetry simplifies the calculation process, especially in cases of algebraic simplifications and solving equations.

Sum of Squares Formula

The sum of squares formula is crucial for calculating the sum of the squares of the first \(n\) natural numbers. This formula is expressed as \(\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).
This formula is very effective in simplifying sums involving squared terms. In the given exercise, understanding this formula allows us to transform the expanded form \(k^2 - 4\) into a more solvable equation.
By substituting \(k^2\) with the sum of squares formula, we accurately calculate the aggregate total, making cumbersome calculations with individual terms unnecessary. Think of it as a shortcut that efficiently delivers the total value of squared terms up to a certain \(n\).

Constant Sum

In algebra, a constant sum signifies adding the same number a certain number of times. The formula \(\sum_{k=1}^{n} c = cn\) applies, where \(c\) is a constant being summed from 1 to \(n\).
For example, in our exercise, \(\sum_{k=1}^{n} 4\) results in \(4n\), which simply adds the value 4 a total of \(n\) times.
This aspect of the solution separates constant elements in an expression for ease of calculation. It simplifies the algebra by isolating constant terms from variable ones, allowing both parts to be summed independently before recombining into a final result.

Expansion of Expressions

Expansion of expressions involves rewriting an algebraic expression in an extended format. This often makes further calculations more straightforward or reveals patterns not initially apparent.
In the original exercise, to expand \((k+2)(k-2)\), knowing it is a difference of squares allows us to convert it into \(k^2 - 4\).
This transformation is foundational in algebra, laying groundwork for solving equations and simplifying expression manipulations. Expansion helps break down complex problems into smaller, more manageable parts, simplifying both solutions and overall understanding of the problem at hand.