Question

Prove by induction for \(n \geq 0\) : $$2+4+6+\cdots+2 n=n^{2}+n$$

Short answer

The statement is proven true for all \( n \geq 0 \) by induction.

Step-by-step solution

Base Case Verification

To start with induction, check if the statement holds for the base case, which is typically the smallest value of \( n \). In this problem, that's \( n = 0 \). For \( n = 0 \), the left side of the equation is the sum of zero terms, which equals 0. The right side of the equation is \( 0^2 + 0 = 0 \). Since both sides are equal, the base case holds.

Inductive Hypothesis

Assume that the statement is true for some arbitrary non-negative integer \( k \), such that the sum of the first \( k \) even numbers is \( 2 + 4 + 6 + \cdots + 2k = k^2 + k \). This assumption is called the inductive hypothesis.

Inductive Step

Now, prove that if the statement holds for \( n = k \), then it must also be true for \( n = k + 1 \). Start with the sum for \( n = k + 1 \): \[2 + 4 + 6 + \cdots + 2k + 2(k+1)\]According to the inductive hypothesis, we know:\[2 + 4 + 6 + \cdots + 2k = k^2 + k\]Adding \( 2(k+1) \) to both sides gives:\[k^2 + k + 2(k+1)\]

Simplify the Expression

Simplify the expression from the inductive step:\[k^2 + k + 2(k+1) = k^2 + k + 2k + 2 = k^2 + 3k + 2\]This can be arranged as follows:\[= (k+1)^2 + (k+1)\]Therefore, the statement holds for \( n = k + 1 \) as well.

Conclusion

By mathematical induction, since the base case holds and if the statement is true for \( n = k \) it implies the truth of the statement for \( n = k + 1 \), the original statement is proven for all \( n \geq 0 \).

Key concepts

Base Case

When using mathematical induction, you first check the smallest possible value of your statement, known as the "base case." For this exercise, our base case is when \( n = 0 \). It is like testing if the approach works right from the start. You calculate both sides of the equation.
On the left side, we have the sum of even numbers which, for \( n = 0 \), equates to nothing, or \( 0 \). The right side gives us \( 0^2 + 0 = 0 \). Both sides are equal, verifying the accuracy for the smallest case. If this had not held, we would immediately identify an error in the statement. Thus, ensuring our base case is correct is crucial in laying the groundwork for the entire induction process.

Inductive Hypothesis

The inductive hypothesis involves making an assumption for some general case \( n = k \). Assume the statement is true for this arbitrary integer \( k \). Here, it means that the sum of the first \( k \) even numbers equals \( k^2 + k \). A hypothesis acts like a stepping stone: we take this as a temporary truth to build upon.
This doesn't mean we've proven the statement for all numbers \( k \), but within the hypothesis, we treat it as if it is correct. By assuming it works for \( n = k \), we can then use this foundation to try and show it holds for \( n = k+1 \). The hypothesis acts like a bridge to the next step.

Inductive Step

After hypothesizing, the next step in mathematical induction is the inductive step. It involves showing that if the statement holds when \( n = k \), it must also hold for \( n = k+1 \). This step is crucial because it attempts to generalize from one specific case to the next.
In our example, we add the next even number, \( 2(k+1) \), to both sides of the equation established in our hypothesis: \( k^2 + k + 2(k+1) \). Simplifying this gives \( (k+1)^2 + (k+1) \). So, if it works for \( k \), it works for \( k+1 \), completing the chain. Essentially, this step shows that our logic process continues beyond \( k \). By proving one case paves the way for the next, we aim for a domino effect making the statement true for each \( n \) that follows.

Proof by Induction

Proof by induction is a cleverly structured method of mathematical proof. It establishes that a statement holds for all natural numbers by first confirming it for the starting point (the base case) and then showing it is true for every subsequent case via the inductive step.
Here's why it’s powerful: Mathematical induction allows you to demonstrate an infinite sequence of truths using a finite amount of logical reasoning. By combining the base case and the solidity of the inductive hypothesis leading to the inductive step, the statement's validity cascades through all natural numbers. This method is like setting up a line of dominos: once the first domino falls (base case), all others (subsequent natural numbers) must follow. For the exercise, with base case verified and an inductive step proved, the statement \( 2+4+6+\cdots+2n=n^{2}+n \) is established."