Prove the odd-number sum with a complete induction step

Define the claim before relying on an ellipsis

The claim is that for every positive integer n, the first n positive odd numbers have sum S(n)=1+3+…+(2n−1)=n². Here n is the number of terms and determines the last term. It does not mean adding odd numbers only up to the numerical value n.

Testing n=1 gives 1, n=2 gives 4, and n=3 gives 9. This suggests a pattern but covers only three inputs. Even testing through 100 does not explain why101 and every larger value must work. Examples can expose errors without proving an infinite claim.

MIT’s open-course induction notes, pages 3–4, state how a base P(1) and a conditional step P(n)⇒P(n+1) for arbitrary positive n establish the claim for all positive integers. The odd-sum proof and repair exercises below are original teaching derivations, not copied lecture examples.

Write the base, assumption and transfer separately

Base: n=1 gives 1 on the left and1²=1 on the right, so P(1) holds. This establishes a genuine starting point rather than treating the first case as an optional obvious detail.

Step: take arbitrary k≥1 and assume P(k), namely S(k)=k². The task is to prove P(k+1) under that assumption. We are not assuming the entire statement for all n, and we are not checking only a particular k.

The kth odd term is2k−1. The next is2(k+1)−1=2k+1. Thus S(k+1)=S(k)+(2k+1)=k²+2k+1=(k+1)². The second equality uses the induction hypothesis; the final equality expands a square.

Because k was arbitrary, the conditional step covers every link. Combined with the established P(1), the induction principle gives S(n)=n² for every n≥1. Identifying arbitrary k is the crucial move beyond finite checking.

Claim: for n≥1, S(n)=n².
Base: S(1)=1=1².
Take arbitrary k≥1 and assume S(k)=k².
Next term:2(k+1)−1=2k+1.
S(k+1)=S(k)+(2k+1)=k²+2k+1=(k+1)².
Therefore P(k)⇒P(k+1). Together with the base, induction establishes all n≥1.

Writing 2k+3 skips an odd number

A common indexing mistake adds an extra2 and writes2k+3 for the next term. At k=2, the first two terms are1 and3, so the next is5;2k+3 instead gives 7. This small check detects the mistake, but the repair still needs the general expression2(k+1)−1.

A valid transfer can still have no starting point

Consider Q(n): n≥5, and the false claim that it holds for all n≥1. If Q(k) holds, then k≥5, so k+1≥5 and Q(k+1) holds. The conditional step is valid. But Q(1) is false, so the statement cannot be concluded for every positive integer.

A transfer rule does not provide its own starting case. Dominoes are only a metaphor: the first must fall and every successive link must transmit the effect. A photograph or a handful of trials is not a mathematical proof.

For the logic of examples versus universal statements, continue with universal claims, existence and counterexamples. Induction additionally needs the positive transfer argument; failing to find a counterexample is insufficient.

Try the even-number sum, then check the complete answer

Transfer exercise: for n≥1, prove 2+4+…+2n=n(n+1). Write the base, arbitrary-k assumption and next term before reading on. If you can only repeat the odd-sum wording, identify the source of every term in the new expression.

Complete reference answer: at n=1, the sum is2 and1×2=2. Take arbitrary k≥1 and assume the first k terms sum to k(k+1). Add the next term2(k+1). The new sum is k(k+1)+2(k+1)=(k+1)(k+2), exactly the target after replacing n with k+1. The base and step establish the result for all n≥1.

Explain why k+1 is the common factor and why the next target is(k+1)(k+2). The self-explanation method for worked examples helps rehearse those reasons. Completion means explaining the steps, not merely recalling the final identity.