Build and check mathematical induction arguments

What an induction proof establishes

Mathematical induction proves statements about successive integers. First verify a starting case P(n₀). Then show, for an arbitrary integer k ≥ n₀, that if P(k) holds, P(k + 1) follows. The base gives the first true case; the implication gives the next one, and repeats to reach any integer in the stated range. This is a proof principle, not an assumption that the final claim is already true

1. 1State a proposition and its integer domain Name the statement P(n), specify the integer starting index n₀ and keep the domain visible. Ordinary induction reaches successive integers beginning at that index, not all real values

Define the sum and the claim

The notation Σ means add the terms obtained by letting the integer index run from its lower bound to its upper bound. Define Sₙ as the sum of 4j − 1 for j = 1, …, n, where n is a positive integer. The first terms are 3, 7 and 11. Our claim P(n) is Sₙ = n(2n + 1). Seeing a pattern in a few sums suggests the claim; induction will justify every positive-integer case

Verify the base case explicitly

The claim starts at n = 1. The sum then has one term, 4 × 1 − 1 = 3. The proposed right side is 1(2 × 1 + 1) = 3. Both sides agree, so P(1) is true. We have checked the base rather than assuming it; we have not yet proved the general step

1. 1Verify the base rather than assume it Substitute n₀ into the claim and directly check both sides or the required property. A statement that the base is true, without the calculation or reason, is not a verification

Which line actually verifies the base case of this sum identity?

State a conditional hypothesis, not the whole theorem

Let k be an arbitrary integer with k ≥ 1 and assume P(k): Sₖ = k(2k + 1). Under this temporary hypothesis, our job is to establish P(k + 1). This proves an implication. It does not assume that the formula holds for every positive integer, and it does not assume the next case. Choosing only k = 3 would not prove the implication for every allowed k.

1. 1Choose an arbitrary admissible index Let k be an arbitrary integer with k ≥ n₀. Temporarily assume P(k). You are proving an implication from that case to the next, not assuming the theorem for every index

Which statement is the induction hypothesis for this proof?

Add the correct next term

The (k + 1)th term is 4(k + 1) − 1 = 4k + 3. Thus Sₖ₊₁ equals Sₖ plus 4k + 3. The target right side is obtained by replacing n everywhere with k + 1, giving (k + 1)(2k + 3). Keep the known k formula and the desired k + 1 formula distinct

1. 1Identify the exact next-case target Replace every occurrence of n by k + 1 when writing P(k + 1). For a finite sum, add the term whose index is k + 1, not the old kth term

Which new term is added when the upper index changes from k to k + 1?

Complete the partial-sum induction proof

The base P(1) was verified. Let k ≥ 1 be an arbitrary integer and assume Sₖ = k(2k + 1). Add the next term and use that hypothesis, as shown below. The resulting factorisation is exactly P(k + 1). We have therefore proved the required implication for every k ≥ 1. By mathematical induction, Sₙ = n(2n + 1) for every integer n ≥ 1.

1. 1Use the hypothesis in a forward derivation Start with the next-case expression or a justified fact, invoke P(k) where appropriate, and derive the target. Do not assume P(k + 1) and merely repeat it
2. 2Distinguish complete proofs from selected steps Successful examples or correct selected steps are not independent construction of a proof. Write the base, arbitrary-index hypothesis, justified next case and quantified conclusion yourself; compare with the ungraded models

Why does the last factorisation match the next-case target?

Written practice 1: a changed finite sum (ungraded)

For every positive integer n, prove by induction that the sum of 2j + 1 from j = 1 to n is n(n + 2). Write the verified base, an arbitrary-index hypothesis, the next term, the justified next-case algebra and the final domain

Model 1: a complete changed-sum proof

At n = 1, the single term is 3 and 1(1 + 2) = 3. Now let k ≥ 1 be an arbitrary integer and assume the sum through k is k(k + 2). The next term is 2(k + 1) + 1 = 2k + 3. Adding it gives the expression below, equal to the claimed formula at k + 1. Thus the next case follows whenever the k case holds. By induction, the identity holds for every positive integer n.

A divisibility claim also has a base and hypothesis

We will prove the displayed divisibility claim for every nonnegative integer. The base calculation is shown below. Let k be an arbitrary nonnegative integer and assume the claim at k. By the definition of divisibility, write the hypothesis using an integer multiplier r, as shown below. The integer condition is essential

Which form correctly expresses the divisibility hypothesis?

Complete the divisibility proof

The base at 0 is verified. Under the hypothesis 5^k − 1 = 4r with integer r, rewrite the next expression as 5(5^k − 1) + 4. Substituting gives 4(5r + 1). Since 5r + 1 is an integer, 4 divides 5^(k+1) − 1. The implication holds for every integer k ≥ 0. By induction, 4 divides 5^n − 1 for every integer n ≥ 0.

1. 1Justify order and divisibility steps A positive multiplier preserves an inequality; a negative multiplier reverses it. Divisibility requires an integer multiplier, and any extra comparison must follow from the stated domain

What completes the divisibility step after obtaining 4(5r + 1), where r is an integer?

An inequality needs a checked base too

Consider 3^n ≥ 2n + 1 for every integer n ≥ 0. At n = 0, both sides are 1, so the weak inequality holds. Let k ≥ 0 be arbitrary and assume 3^k ≥ 2k + 1. The target is 3^(k+1) ≥ 2k + 3. We must justify how the hypothesis leads to that target, not merely replace k by k + 1 in the assumption

What follows on multiplying the inequality hypothesis by 3?

Complete the inequality induction proof

The base at 0 is true. Let k ≥ 0 be an arbitrary integer and assume 3^k ≥ 2k + 1. Multiply by positive 3 to obtain 3^(k+1) ≥ 6k + 3. Since k ≥ 0, the difference (6k + 3) − (2k + 3) = 4k is nonnegative. Hence the target inequality at k + 1 follows. By induction, 3^n ≥ 2n + 1 for every integer n ≥ 0. Each comparison has its own reason

For k ≥ 0, which fact justifies the second comparison 6k + 3 ≥ 2k + 3?

Written practice 2: a changed inequality (ungraded)

Prove by induction that 4^n ≥ 3n + 1 for every integer n ≥ 0. Verify the base, state the hypothesis and target, and explain both the positive-multiplier step and the remaining comparison. Write the complete argument before reading its model

Model 2: justify both inequality steps

At n = 0, 4^0 = 1 = 3 × 0 + 1. Let k ≥ 0 be arbitrary and assume 4^k ≥ 3k + 1. Multiplying by positive 4 yields 4^(k+1) ≥ 12k + 4. The difference between 12k + 4 and the target 3k + 4 is 9k ≥ 0. Thus the claim follows at k + 1. By induction, 4^n ≥ 3n + 1 for every nonnegative integer n.

Written practice 3: a changed divisor (ungraded)

Prove by induction that 6 divides 7^n − 1 for every integer n ≥ 0. Use an explicit integer multiplier in the hypothesis and identify a new integer multiplier in the next case. Include the base and final quantified conclusion; no proof score is awarded

Model 3: the new integer witness

At n = 0, 7^0 − 1 = 0 = 6 × 0. Let k ≥ 0 be an arbitrary integer and assume 7^k − 1 = 6r for an integer r. The next expression is 7(7^k − 1) + 6 = 42r + 6 = 6(7r + 1). Since 7r + 1 is an integer, this proves the next divisibility case. By induction, 6 divides 7^n − 1 for every integer n ≥ 0.

Starting indices and step size matter

A proof with a base at 4 and a valid next-integer step for k ≥ 4 establishes the claim only for integers n ≥ 4; earlier cases need separate checking. A step from k to k + 2 has two chains. Starting only at 0 reaches the even indices but not the odd ones. To cover every nonnegative integer with that step, verify both 0 and 1 and justify the step throughout the domain

1. 1Conclude over exactly the verified range After verifying the base and proving P(k) implies P(k + 1) for every allowed k, conclude P(n) for every integer n ≥ n₀. A shifted base does not settle smaller indices. A step of 2 needs both parity chains to cover all integers

With only P(0) checked and a valid step from k to k + 2, which part remains unproved?

Written practice 4: a shifted sum (ungraded)

For every integer n ≥ 2, prove that the sum of 2j − 1 from j = 2 to n is n² − 1. Notice the lower summation bound and the starting index. Verify the actual base, add the correct next term and give a complete induction conclusion

Model 4: respect the shifted starting point

At n = 2, the sum has one term 3, and 2² − 1 = 3. Let k ≥ 2 be an arbitrary integer and assume the sum from j = 2 through k is k² − 1. Add the next term 2(k + 1) − 1 = 2k + 1. The result below is the claimed expression at k + 1. Thus the implication holds for every k ≥ 2; by induction, the identity holds for all integers n ≥ 2.

Audit the logical chain

A true base without a general step is insufficient: P(n) meaning n ≤ 4 is true at 0 but fails to pass from 4 to 5. A valid step without the required base is also insufficient: P(n) meaning n ≥ 1 has a valid next-integer implication on n ≥ 0, yet P(0) is false. Do not assume the target, skip indices, reverse an inequality incorrectly or omit the conclusion. The written models are teaching; selected steps do not establish independently constructed proofs

1. 1Distinguish complete proofs from selected steps Successful examples or correct selected steps are not independent construction of a proof. Write the base, arbitrary-index hypothesis, justified next case and quantified conclusion yourself; compare with the ungraded models

For nonnegative integers, P(n) means n ≥ 1. The implication P(k) ⇒ P(k + 1) is valid. Which required ingredient is missing for a conclusion starting at 0?