Problem 27 - 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 $$\sum_{k=1}^{n} k^{2}=\frac{n(n+1)(2 n+1)}{6}$$ $$ \sum_{k=1}^{n} 4(k-1)^{2} $$

Step-by-Step Solution

Answer

The sum is \( \frac{2n(n+1)(2n-3)}{3} \).

Step 1: Expand the Expression Inside the Sum

Rewrite the given sum by expanding the expression inside it. Start with \( \sum_{k=1}^{n} 4(k-1)^{2} \). Factor out the constant 4, leading to \( 4 \sum_{k=1}^{n} (k-1)^{2} \).

Step 2: Reformulate the Expression Inside the Sum

Recognize that \((k-1) = k - 1\), and rewrite \((k-1)^2 = (k - 1)^2 = k^2 - 2k + 1\). Thus, the sum becomes \( 4 \sum_{k=1}^{n} (k^2 - 2k + 1) \).

Step 3: Distribute the Sum

Distribute the summation over each term: \( 4 \left( \sum_{k=1}^{n} k^2 - 2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \right) \).

Step 4: Use Known Summation Formulas

Substitute the known formulas for each component: \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \), \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \), and \( \sum_{k=1}^{n} 1 = n \).

Step 5: Plug in the Formulas into the Expression

Substitute the summation formulas into the expanded expression: \( 4 \left( \frac{n(n+1)(2n+1)}{6} - 2 \frac{n(n+1)}{2} + n \right) \).

Step 6: Simplify the Expression

Simplify the expression by performing operations: compute each term and combine. The expression becomes \( \frac{2n(n+1)(2n+1)}{3} - 4n(n+1) + 4n \).

Step 7: Combine Like Terms

Combine like terms to further simplify. Rewrite the expression with common denominators when necessary to combine fractions: \( \frac{2n(n+1)(2n+1)}{3} - \frac{12n(n+1)}{3} + \frac{12n}{3} \). Simplifying leads to \( \frac{2n(n+1)(2n-3)}{3} \).

Key Concepts

Summation Formulas

Evaluating algebraic sums efficiently is possible by utilizing various summation formulas. These formulas help condense multiple components into simple expressions. For instance:

• The formula for the sum of the first \( n \) integers is given by \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \), providing a concise way to calculate sequences quickly.
• The formula for the sum of squares of the first \( n \) integers is \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \), useful when dealing with problem that requires evaluating quadratic sums.

These expressions come in handy when breaking down more complex summations because they allow the isolation of principal components. Recognizing patterns and understanding these forms enable simplification at a glance. In exercises, substituting these formulas simplifies otherwise tedious calculations within summations.

Sum of Squares

The sum of squares plays a crucial role in many algebraic computations. This is important when evaluating expressions where each term is squared.
Using the sum of squares formula, \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\), allows for the efficient computation of these terms:

• First, identify that the task is to find the sum of squares.
• Apply the formula as a substitute for the series being evaluated.

The calculation becomes manageable by ignoring each individual squared term and focusing on the overall sum. This approach is particularly helpful for larger sequences, as the formula aggregates all terms succinctly. By understanding this concept, you can transfer this method to similar exercises, simplifying solutions by leveraging pattern recognition and match formulas accordingly.

Algebraic Expansion

Algebraic expansion is an essential skill when dealing with sums, like the given expression \( \sum_{k=1}^{n} 4(k-1)^{2} \). Expanding this algebraically means rewriting it such that the expression inside the sum is explicitly detailed.
When expanding, you:

• First factor out constants, which simplifies each term. For example, extracting the constant 4 from the sum gives \( 4 \sum_{k=1}^{n} (k-1)^2 \).
• Next, expand the expression \( (k-1)^2 \) using identity \( k^2 - 2k + 1 \).
• This turns the problem into evaluating sums of individually straightforward components.

In this exercise, breaking complex terms into simpler ones lets us utilize known summation formulas directly. Each step reduces the complexity level, allowing a focus on arithmetic operations and summation techniques. Mastering expansion bridges the gap between intricate expressions and simpler evaluative methods, vital for advancing in mathematics.