Induction for sums

Objective. Prove finite summation identities.

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Learn: the key idea

Induction for sums proves a formula for Σ (a series of terms) is correct for every positive integer n, by checking a base case and then showing the formula for n=k+1 follows from assuming it for n=k. The key algebraic step adds the (k+1)th term to both sides of the assumed formula for the sum up to k, then factorises to match the target formula with n replaced by k+1. For example, proving Σr from 1 to n equals n(n+1)/2 involves adding (k+1) to k(k+1)/2 and simplifying to (k+1)(k+2)/2.

Worked example

Prove by induction that the sum of the first n cubes, Σr³ for r=1 to n, equals [n(n+1)/2]² for all positive integers n.

1 $Base case n=1: left side = 1, right side = [1\times2/2]² = 1, so true for n=1$
2 $Assume true for n=k: Σr³ = [k(k+1)/2]² up to r=k$
3 $Add (k+1)³: [k(k+1)/2]² + (k+1)³ = (k+1)²[k²/4 + (k+1)] = (k+1)²(k²+4k+4)/4 = [(k+1)(k+2)/2]²$
4 $This matches the formula with n=k+1, so by induction the result holds for all positive integers n$

Answer: Proved: Σr³ = [n(n+1)/2]² by induction

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Common mistake. Students substitute n=k+1 directly into the formula to 'prove' it, rather than building it from the assumed case for n=k plus the next term. The proof must start from the k-case sum, not assume the result being proved.

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