proving summation formulas using induction — HSC mathematics extension 1, live

Proving summation formulas using induction

Checking 1 + 2 + … + n = n(n+1)/2 for n = 1, 2, 3, 4, 5 proves it for five numbers and no more. Induction proves it for every positive integer with two moves: verify n = 1 directly (both sides equal 1), then take an unspecified k, assume the sum to k is k(k+1)/2, and add the next term. The algebra k(k+1)/2 + (k+1) = (k+1)(k+2)/2 is the formula with k + 1 in place of k, so truth at any k forces truth at k + 1, and the base case starts the chain. This is the real knowscape from knowhere, not a picture of one. Drag it. Watch what actually changes.

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why this one carries the topic.

Induction never proves the formula outright — it proves that truth is contagious. Show it holds once, then show that whenever it holds it hands itself to the next case, and the whole infinite chain falls for free. You're not summing anything; you're building a machine that sums forever.

A shaded block stands for the assumed sum S_k = k(k+1)/2 and an amber block for the next term k + 1; the bridge equation beneath shows their total factorising to (k+1)(k+2)/2, while the base case view substitutes n = 1 and both sides read 1.

what examiners catch — The usual slip is to write the formula for k + 1 and simplify both sides until they match, never using the hypothesis. What earns the mark is splitting the sum to k + 1 as the assumed sum to k plus the term (k+1), substituting k(k+1)/2 and factorising to (k+1)(k+2)/2, then stating the conclusion for all positive integers.

five things, not forty.

The base case substitutes n = 1 into both sides and shows they agree: 1 on the left, 1(1+1)/2 = 1 on the right.

The inductive hypothesis is written for an arbitrary k, not a particular number: assume 1 + 2 + … + k = k(k+1)/2.

The inductive step splits the sum to k + 1 as the assumed sum to k plus the new term (k+1), then substitutes the hypothesis.

Factorising k(k+1)/2 + (k+1) gives (k+1)(k+2)/2, the formula with k + 1 in place of k, which completes the step.

Base case plus inductive step together prove the statement for all positive integers; a list of checked values never does.

nothing here is a standalone fact.

knowhere maps every concept to what it rests on and what rests on it — 4 underneath this one, 1 built on top. Each one says why, in a sentence, not as an arrow on a diagram.

this conceptproving summation formulas using inductionA shaded block stands for the assumed sum S_k = k(k+1)/2 and an amber block for the next term k + 1; the bridge equation beneath shows their total factorising to (k+1)(k+2)/2, while the base case view substitutes n = 1 and both sides read 1.

sits under itthe base casebecause no anchor, no induction

sits under itthe inductive stepbecause the step is the engine of the proof

sits under itthe inductive hypothesisbecause The step assumes the formula for an arbitrary k before adding the next term.

sits under itdirect proof and proof by casesbecause Knowing what counts as a proof explains why checking cases is not one.

built on itdiagnosing a false induction proofbecause A few correct proofs make the shape of a broken one easier to see.

5 more, same treatment.

Each one is its own knowscape in the app — built for how a particular student takes things in, not one explanation handed to everybody.

divisibility proofs by inductionin the appthe inductive hypothesisin the appthe inductive stepin the appthe base casein the appdiagnosing a false induction proofin the app

they don't read it. they run it.

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