Question

Prove by induction that 8 divides \((2 n+1)^{2}-1\) for all \(n \in \mathbb{N}\).

Short answer

8 divides \((2n+1)^2 - 1\) for all \(n \in \mathbb{N}\) by induction.

Step-by-step solution

Base Case

First, we'll show that the statement holds for the initial value of \( n = 1 \). Calculate \((2(1) + 1)^2 - 1\):\[(3)^2 - 1 = 9 - 1 = 8\]Since 8 divides 8, the base case holds true.

Inductive Hypothesis

Assume the statement is true for some integer \( k \), i.e., assume that 8 divides \((2k + 1)^2 - 1\). This means:\[(2k + 1)^2 - 1 = 8m\]for some integer \( m \).

Inductive Step

We need to show that if the statement is true for \( n = k \), then it is also true for \( n = k + 1 \). Calculate \((2(k + 1) + 1)^2 - 1\):\[((2k + 2) + 1)^2 - 1 = (2k + 3)^2 - 1\]Expand the square:\[(2k + 3)^2 = (2k + 3)(2k + 3) = 4k^2 + 12k + 9\]Subtract 1:\[4k^2 + 12k + 9 - 1 = 4k^2 + 12k + 8\]This expression can be factored as:\[4k(k + 3) + 8\]Notice \( 8 \) can be factored out:\[8(k(k + 3)/2 + 1)\]Thus, \( (2(k + 1) + 1)^2 - 1 \) is divisible by 8.

Conclusion

Since the base case holds and the inductive step shows that if the statement is true for \( n = k \) then it is true for \( n = k + 1 \), by mathematical induction, the statement is true for all \( n \in \mathbb{N} \).

Key concepts

Base Case

In mathematical induction, the Base Case is the initial step that sets the foundation for proving a statement is true for all natural numbers. Think of it as the first step on a ladder. If you stabilize the first step, the next ones can follow safely. For the given problem, we start by checking that the expression \((2n+1)^2 - 1\) is divisible by 8 when \(n=1\).

First, substitute \(n = 1\) into the formula to calculate \((2(1) + 1)^2 - 1\):
\[\begin{align*} (3)^2 - 1 &= 9 - 1 \ &= 8 \end{align*}\]
This calculation shows that the result, 8, is indeed divisible by itself. Hence, the base case holds true. Validating the base case is like securing the first rung of a ladder, ensuring that the climb ahead is possible.

Inductive Hypothesis

The Inductive Hypothesis is the "assume step" in mathematical induction where you assume the statement holds for a particular integer, say \(n = k\). This helps bridge the gap to prove the next step. Think of it as hypothesizing the sturdiness of a step at height k.

In our scenario, assume the statement is true for \(n = k\):
Assume \(8\) divides \((2k + 1)^2 - 1\). This implies:
\[(2k + 1)^2 - 1 = 8m\]
for some integer \(m\). Essentially, this step presumes that from the base to the current height \(k\), our ladder is strong.

This hypothesis does not require proving at this moment. It's a strategic assumption to prove the possibility for the next step which is the inductive step.

Inductive Step

The Inductive Step is the process of proving the statement for \(n = k + 1\), based on the assumption that it holds for \(n = k\). It's like ensuring every step beyond the current secure rung, all the way up the ladder, will also be secure.

We begin with replacing \(n\) with \(k + 1\) in the original expression:
Calculate \((2(k + 1) + 1)^2 - 1\):
\[\begin{align*} (2(k + 1) + 1)^2 - 1 &= (2k + 3)^2 - 1 \ &= (2k + 3)(2k + 3) - 1 \ &= 4k^2 + 12k + 9 - 1 \ &= 4k^2 + 12k + 8 \end{align*}\]
This result can be written as:
\[4k(k+3) + 8 = 8(k(k+3)/2 + 1)\]
Here, notice that 8 factors out of the entire expression, which confirms that for \(n = k + 1\), the stated expression is divisible by 8.

This reasoning completes the inductive step. If the base case is true (it was), and if reaching the next step is possible assuming previous strength (proved here), the statement must hold true for all \(n \in \mathbb{N}\).