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Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ 2+7+12+\cdots+(5 n-3)=\frac{1}{2} n(5 n-1) $$

Short Answer

Expert verified
The statement is true for all positive integers \(n\) by mathematical induction.

Step by step solution

01

Base Case

Start by verifying the statement for the smallest positive integer, which is \(n = 1\). Substitute \(n = 1\) into both sides of the equation:- Left-hand side (LHS): \(5 \times 1 - 3 = 2\).- Right-hand side (RHS): \(\frac{1}{2} \times 1 \times (5 \times 1 - 1) = \frac{1}{2} \times 1 \times (5 - 1) = \frac{1}{2} \times 4 = 2\).Since LHS equals RHS when \(n = 1\), the base case holds true.
02

Inductive Hypothesis

Assume that the statement is true for some positive integer \(k\), i.e.,\[2 + 7 + 12 + \cdots + (5k - 3) = \frac{1}{2} k(5k - 1).\]This is the inductive hypothesis.
03

Inductive Step

Show that the statement is true for \(n = k+1\), meaning you need to prove:\[2 + 7 + 12 + \cdots + (5k-3) + (5(k+1) - 3) = \frac{1}{2} (k+1)(5(k+1) - 1).\]Start with the left-hand side using the hypothesis:\[(2 + 7 + 12 + \cdots + (5k-3)) + (5(k+1) - 3) = \frac{1}{2} k(5k-1) + (5(k+1) - 3).\]
04

Simplification

Simplify the expression for \(5(k+1) - 3\):\[5(k+1) - 3 = 5k + 5 - 3 = 5k + 2.\]Now plug this back into the LHS of our working equation:\[\frac{1}{2} k(5k-1) + (5k + 2).\]Simplify inside the brackets:\[\frac{1}{2}(5k^2 - k) + 5k + 2.\]
05

Combine and Adjust Terms

Factor and simplify:- Combine \(\frac{1}{2}(5k^2 - k)\) with the terms \(5k + 2\):\[= \frac{1}{2}(5k^2 - k) + \frac{10k}{2} + \frac{4}{2}.\]- Simplify to obtain:\[= \frac{1}{2}(5k^2 - k + 10k + 4).\]}
06

Verification and Final Formula

Collect like terms inside the brackets:\[\frac{1}{2}(5k^2 + 9k + 4).\]Notice it matches the intended expression for \(k+1\):\[\frac{1}{2}(k+1)(5(k+1) - 1)\]Rewriting gives:\[(5(k+1)^2 - (k+1)) = 5(k+1)^2 - (k + 1).\]Thus proving the formula holds for \(n = k+1\). Hence, by induction, it is true for all positive integers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductive Hypothesis
The inductive hypothesis is a crucial part of mathematical induction, a technique used to prove that a statement is true for all positive integers. This concept involves assuming that a statement is true for a particular positive integer, often denoted as \(k\).
We use this assumption to solve a step further. The inductive hypothesis for our example assumes that the formula \(2 + 7 + 12 + \cdots + (5k - 3) = \frac{1}{2} k(5k - 1)\) holds true for some integer \(k\). This assumption acts as a bridge to prove the next step.
By assuming the statement is true for \(k\), we can investigate whether this supposition allows the statement to also hold for \(k+1\), thereby reinforcing the pattern or formula of the entire sequence. This hypothesis serves as the foundation for completing the proof that the statement applies to all subsequent integers.
Base Case
The base case is the starting point of any induction proof. It establishes the initial truth of the statement for the smallest positive integer in your series, commonly \(n = 1\).
To validate our statement at the base case: for \(n = 1\), the left-hand side evaluates to \(2\), since \(5 \times 1 - 3 = 2\).
The right-hand side also simplifies to \(2\), as \(\frac{1}{2} \times 1 \times (5 \times 1 - 1) = 2\).
  • Verify by substituting: This simple test ensures that at the beginning of the sequence, both expressions equate to the same result.
Verifying the base case confirms that our starting point holds true, laying the groundwork for the subsequent step-by-step induction process.
Inductive Step
The inductive step is where we show that if the statement holds for some positive integer \(k\), then it must hold for \(k+1\). Here, inductive reasoning is fully in action.
For our problem, once the statement is assumed true for \(k\), we need to demonstrate it for \(k+1\). Begin by taking the expression \((2 + 7 + 12 + \cdots + (5k - 3)) + (5(k+1) - 3)\).
Using the inductive hypothesis, substitute in the known formula for the sequence up to \(k\), then simplify and combine:
  • Replace terms: First, implement the step using the inductive assumption into your current equation.
  • Simplify: Work to reduce it down to show how it will hold for \(k+1\).
This step involves algebraic manipulation to demonstrate the next iteration, ensuring that the initial inductive hypothesis is what enables this future proofing.
Algebraic Proof
Algebraic proof in the context of mathematical induction involves a careful blend of hypothesis and algebra skills to establish a statement's validity across a sequence. This typically includes steps where you:
1. Substitute: Start from where left off—the inductive hypothesis provides a known form.
2. Simplify: Perform algebraic manipulation like combining terms and expanding brackets.
3. Compare: Each term should align with the desired form for \(k+1\).
You must ensure that terms are combined correctly and transform the expression into the exact form of the right side for \(n = k+1\).
This technique concludes with showing that our simplified form and the expected formula for \(k+1\) are identical, thereby completing the proof that the statement holds for all \(n\), fulfilling our algebraic proof requirements.

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