Problem 3 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers n. $$5+8+11+\cdots+(3 n+2)=\frac{n(3 n+7)}{2}$$

Step-by-Step Solution

Answer

The formula is valid for all natural numbers using mathematical induction.

Step 1: Base Case

To begin, verify the formula for the smallest natural number, which is usually \( n = 1 \). Substitute 1 into both sides of the formula. The left side becomes: \( 3(1) + 2 = 5 \).The right side becomes: \( \frac{1(3(1) + 7)}{2} = \frac{1(10)}{2} = 5 \).Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume the formula is valid for some arbitrary natural number \( k \). This means we assume that\[ 5 + 8 + 11 + \, \cdots \, + (3k+2) = \frac{k(3k+7)}{2}. \]This assumption is called the inductive hypothesis.

Step 3: Inductive Step

Show that if the formula holds for \( n = k \), then it also holds for \( n = k+1 \).First, consider the sum up to \( n = k+1 \):\[ 5 + 8 + 11 + \, \cdots \, + (3k+2) + (3(k+1)+2). \]Using the inductive hypothesis, the part up to \( (3k+2) \) is \( \frac{k(3k+7)}{2} \),so the sum simplifies to:\[ \frac{k(3k+7)}{2} + (3k + 5). \]

Step 4: Algebraic Simplification

Simplify the expression from Step 3 to see if it equals \( \frac{(k+1)(3(k+1)+7)}{2} \).Start by simplifying the expression:\[ \frac{k(3k+7)}{2} + 3k + 5 = \frac{k(3k+7) + 2(3k + 5)}{2}. \]Expand:\[ \frac{3k^2 + 7k + 6k + 10}{2} = \frac{3k^2 + 13k + 10}{2}. \]Now, simplify the right side of the desired equation:\[ \frac{(k+1)(3k+3+7)}{2} = \frac{(k+1)(3k+10)}{2}. \]Expand:\[ \frac{k \cdot 3k + 10k + 3k + 10}{2} = \frac{3k^2 + 13k + 10}{2}. \]Notice both expressions are identical.

Step 5: Conclusion

Since the inductive step holds, we conclude by induction that the formula is true for all natural numbers \( n \). Thus, mathematical induction shows that the given expression is valid for all natural numbers.

Key Concepts

Base Case

In mathematical induction, the base case serves as the starting point. It involves proving that the statement or formula is true for the initial value of the series, usually for the smallest natural number, like 1. In our problem, we begin by substituting \( n = 1 \) into both sides of the equation. This means calculating the sum on the left and comparing it to the formula's right side expression. Here's a quick look:

• Left Side: Substitute \( n = 1 \), giving us \( 5 \) because it simplifies to \( 3(1) + 2 \).
• Right Side: Substitute \( n = 1 \) into \( \frac{n(3n+7)}{2} \), yielding \( \frac{1(10)}{2} = 5 \).

Both sides are equal, which confirms our base case holds true. Establishing this equality is crucial; it lays down the foundation for the inductive step that follows. Without the base case, the structure of induction couldn't proceed.

Inductive Hypothesis

The inductive hypothesis is at the heart of mathematical induction. It is where we "assume" our formula is correct for a certain natural number, usually denoted as \( k \). This assumption allows us to build a bridge that links the base case to the rest of the numbers. In our scenario, we assume that the formula:\[5 + 8 + 11 + \, \cdots \, + (3k+2) = \frac{k(3k+7)}{2}.\]is true for \( n = k \). This step doesn't involve proving anything on its own, but it is a necessary component.Think of the inductive hypothesis as a stepping stone. Assume it's a truth we hold for the higher ground we need to reach in proving the next point in the sequence. Without this assumption, we couldn't logically progress to \( n = k + 1 \).

Inductive Step

The inductive step is where the magic of induction happens. We take our inductive hypothesis and aim to prove that if it's true for \( n = k \), then it must also be true for \( n = k+1 \). This step not only validates our assumption but propels the induction forward.To illustrate:

• Start by expressing the sequence sum up to \( n = k+1 \): \( 5 + 8 + 11 + \, \cdots \, + (3k+2) + (3(k+1)+2) \).
• Using the inductive hypothesis, the sum up to \( 3k+2 \) equals \( \frac{k(3k+7)}{2} \).
• Add the next term \( (3k + 5) \) to this sum.

What we do here is break the sum into parts we know by our hypothesis and the new elements.Next, we simplify the expression:\[\frac{k(3k+7)}{2} + 3k + 5 = \frac{(k+1)(3(k+1)+7)}{2}\]This algebraic step proves both sides are equal, indicating that if it's true for \( n = k \), it holds for \( n = k+1 \) as well.Successfully completing the inductive step affirms that our formula works for all natural numbers, seamlessly connecting them, thanks to induction. This powerful technique constructs a comprehensive proof one small, logical step at a time.