fact_odd_sum_main
For every integer $n \ge 1$, the sum of the first $n$ positive odd numbers equals $n^2$; that is, $S(n) = n^2$, where $S(n) = 1 + 3 + 5 + \cdots + (2n-1)$.
$ npx -y skills add frenzymath/Danus --agent claude-codeHow it fires
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- Fires itselfAuto-invocation. Claude auto-loads it when your prompt matches the work.Auto-invocation is when the right skill fires by itself at the right moment, driven by a FLOW.md router and a hook, instead of you invoking it by name. It is the difference between a skill being installed and a skill actually getting used.Read the full definition →
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Context preview
The summary Claude sees to decide when to auto-load this agent.
For every integer $n \ge 1$, the sum of the first $n$ positive odd numbers equals $n^2$; that is, $S(n) = n^2$, where $S(n) = 1 + 3 + 5 + \cdots + (2n-1)$.
Agent definition
fact_odd_sum_main.mdfact_id: fact_odd_sum_main
problem_id: odd-sum
author: example-worker
predecessors: [fact_odd_recurrence, fact_square_recurrence]
glossary_introduces: {}
external_refs: [{"key": "Exm20", "authors": ["C. Example"], "title": "Elementary induction, revisited", "venue": "Example Lecture Notes", "year": "2020", "cited_for": "the principle of mathematical induction in the form used here"}]statement
For every integer $n \ge 1$, the sum of the first $n$ positive odd numbers equals $n^2$; that is, $S(n) = n^2$, where $S(n) = 1 + 3 + 5 + \cdots + (2n-1)$.
proof
We argue by induction on $n$, in the form recalled in \cite{Exm20}.
Base case. For $n = 1$ we have $S(1) = 1 = 1^2$.
Inductive step. Suppose $S(n) = n^2$ for some $n \ge 1$. By the recurrence for the partial sums, $S(n+1) = S(n) + (2n+1)$. Substituting the inductive hypothesis gives $S(n+1) = n^2 + (2n+1)$. By the recurrence for the squares, $n^2 + (2n+1) = (n+1)^2$. Hence $S(n+1) = (n+1)^2$, completing the induction.
intuition
Both $S(n)$ and $n^2$ start at $1$ and grow by the same increment $2n+1$ at each step, so they agree for all $n$. The induction simply records that two sequences with equal initial value and equal one-step increments coincide.
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