Question

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1+5+9+\cdots+(4 n-3)=n(2 n-1) $$

Short answer

By induction, \(1 + 5 + 9 + \cdots + (4n-3) = n(2n-1)\) is true for all natural numbers \(n\).

Step-by-step solution

Base Case

Verify the given statement for the first natural number, which is typically \(n = 1\). Substitute \(n = 1\) into both sides of the given equation and check for equality:Left-hand side: \begin{align*}4(1) - 3 &= 1enRight-hand side: n(2n - 1) = 1(2(1) - 1) = 1enThus, both sides are equal, confirming that the statement holds true for the base case \(n = 1\).

Inductive Hypothesis

Assume the statement is true for some arbitrary natural number \(k\). This means:\begin{align*}1 + 5 + 9 + \cdots + (4k - 3) = k(2k - 1)

Inductive Step

Using the inductive hypothesis, the goal is to show that the statement also holds for \(k + 1\). Consider the sum for \(k+1\) terms:\begin{align*}1 + 5 + 9 + \cdots + (4k - 3) + [4(k+1) - 3]By the inductive hypothesis, we know:1 + 5 + 9 + \cdots + (4k - 3) = k(2k - 1)So,k(2k - 1) + [4(k+1) - 3]Simplify the expression:k(2k - 1) + [4k + 4 - 3] = k(2k - 1) + 4k + 1 =k(2k - 1) +4k + 1 eq=2k^2 - k + 4k + 1 eq= 2k^2 + 3k + 1 eq= (k + 1)(2(k + 1) - 1)eq= (k + 1)(2k + 2 - 1) eq= (k + 1)(2k + 1)qeeq This confirms that the expression is also true for \(n = k + 1\).

Conclusion

Since the base case is true and the inductive step has been proven, by the Principle of Mathematical Induction, the statement \(1+5+9+ \cdots+(4n-3)=n(2n-1)\) is true for all natural numbers \(n\).

Key concepts

Base Case

The base case is the starting point in the process of Mathematical Induction. It involves proving that the statement holds true for the first natural number, typically when \( n = 1 \). In our exercise, we start by substituting \( n = 1 \) into the equation:

Left-hand side: \( 4(1) - 3 = 1 \)
Right-hand side: \( 1(2(1) - 1) = 1 \).

Since both sides of the equation are equal, the base case holds true. This is a crucial step as it establishes the groundwork for the entire induction process. Without verifying the base case, we cannot proceed with the induction.

Inductive Hypothesis

The inductive hypothesis assumes that the statement is true for an arbitrary natural number \( k \). This assumption is necessary for the next step, which aims to prove the statement for \( k+1 \).

For our exercise, the inductive hypothesis assumes:
\(1 + 5 + 9 + \cdots + (4k - 3) = k(2k - 1)\).
This assumes that the sum of the first \( k \) terms of our sequence equals \( k(2k - 1) \). While this might seem like a leap of faith, it's a necessary assumption to bridge the gap to the next step in our proof.

Inductive Step

The inductive step builds on the inductive hypothesis. Here, we must prove that if the statement holds for \( k \), then it also holds for \( k+1 \). This step involves some algebraic manipulation to extend the assumption to the next case.

Start with the sum up to \( k+1 \):
\( 1 + 5 + 9 + \cdots + (4k - 3) + [4(k+1) - 3] \).
By the inductive hypothesis, this becomes:
\( k(2k - 1) + [4(k+1) - 3] \).

Simplify it:
\( k(2k - 1) + [4k + 4 - 3] = k(2k - 1) + 4k + 1 \).

Further simplify to:
\( 2k^2 - k + 4k + 1 = 2k^2 + 3k + 1 \).

Finally, factorize to match the form of our original statement:
\( (k + 1)(2k + 1) \).

This completes the inductive step, showing that the statement holds for \( k+1 \) if it holds for \( k \).

Summation Formula

A summation formula provides a simplified expression for the sum of a sequence of numbers. In our exercise, the summation formula is:
\( 1 + 5 + 9 + \cdots + (4n - 3) = n(2n - 1) \).

This formula lets us compute the sum directly without adding each term individually. The process of mathematical induction effectively verifies that this summation formula is correct.

By using induction:

• Start with a base case (\( n = 1 \)).
• Assume it holds for \( k \) (inductive hypothesis).
• Prove it for \( k + 1 \) (inductive step).

We confirm that the given summation formula is valid for all natural numbers \( n \). This method shows the power and elegance of mathematical induction in proving summation formulas and other mathematical statements.