CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ , and $\sum_{k < n + 1} (- 1)^{k} ι ⁣ (\binom{n}{k}) = 0$ for $n \geq 1$

Statement

The row sum. For every $n \in N$ , in $N$ , $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n} .$
The alternating row sum. For every $n \geq 1$ , in $R$ , $\sum_{k < n + 1} (- 1)^{k} ι (\binom{n}{k}) = 0.$

Clause 2 is false at $n = 0$ , where the sum has the single term $(- 1)^{0} ι (\binom{0}{0}) = 1$ . The hypothesis $n \geq 1$ is therefore part of the statement, and this page's companion records the version that drops it as a false statement.

Facts & Assumptions

Given: A natural $n$ , a finite set $A$ with $∣ A ∣ = n$ , and $ι$ the canonical natural (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L1]

The binomial theorem: $(x + y)^{n} = \sum_{k < σ (n)} ι (\binom{n}{k}) x^{k} y^{n - k}$ for reals $x$ , $y$ (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ ).

[L2]

$∣ P (A) ∣ = 2^{n}$ ( $∣ P (A) ∣ = 2^{∣ A ∣}$ for finite $A$ ), and $∣ [A]^{k} ∣ = (\binom{n}{k})$ , with $[A]^{k} = \emptyset$ for $k > n$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L3]

The sum rule for a partition indexed by a finite set (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition), the bridge $\sum_{i \in N} u_{i} = \sum_{k < N} u_{k}$ for an index set that is a natural number (The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form), and the $N$ -valued recursion clauses (Finite sums and finite products of natural numbers, $\sum_{k < n} a_{k}$ and $\prod_{k < n} a_{k}$ in $N$ ).

[L4]

$ι$ is additive, multiplicative and injective, and $ι (\sum_{k < N}^{N} a_{k}) = \sum_{k < N}^{R} ι (a_{k})$ (clauses 0, 6 and 7 of Laws of finite sums and products in $N$ , and $ι (\sum_{k < n} a_{k}) = \sum_{k < n} ι (a_{k})$ ).

[L5]

Powers: $ι (m^{j}) = ι (m)^{j}$ (Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ , clause (d)); $a^{0} = 1$ and $a^{σ (j)} = a^{j} a$ , so $1^{j} = 1$ and $0^{j} = 0$ for $j \geq 1$ (Integer powers $a^{m}$ , Multiplication by zero: $0 \cdot a = 0$ , Field).

[L6]

Real finite sums: the recursion clauses and the scaling clause (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L7]

Subsets of a finite set are finite and have cardinality at most $n$ (A subset of a finite set is finite, with $∣ B ∣ \leq ∣ A ∣$ , and equality holds if and only if $B = A$ , The cardinality $∣ A ∣$ of a finite set); induction (The principle of mathematical induction).

Proof

technique · direct

1.1

Clause 1, by counting. Every $S \subseteq A$ satisfies $∣ S ∣ \leq n$ by [L7], so $P (A)$ is the union of the sets $[A]^{k}$ for $k \in σ (n) = {0, 1, \dots, n}$ ; these are pairwise disjoint, since $S$ determines $∣ S ∣$ . By [L3] and [L2], $2^{n} = ∣ P (A) ∣ = \sum_{k \in σ (n)} ∣ [A]^{k} ∣ = \sum_{k < σ (n)} (\binom{n}{k})$ , the last equality being the bridge for an index set that is a natural number.

L2L3L7

1.2

Clause 2. Apply [L1] with $x = - 1$ and $y = 1$ . The left-hand side is $(- 1 + 1)^{n} = 0^{n}$ , which is $0$ because $n \geq 1$ ([L5]). The right-hand side is $\sum_{k < σ (n)} ι (\binom{n}{k}) (- 1)^{k} 1^{n - k} = \sum_{k < σ (n)} (- 1)^{k} ι (\binom{n}{k})$ , using $1^{j} = 1$ from [L5]. Hence the alternating sum is $0$ .

L1L5

2.1

Clause 1 again, from the binomial theorem, as a check that the two routes agree. Taking $x = y = 1$ in [L1] gives $(1 + 1)^{n} = \sum_{k < σ (n)} ι (\binom{n}{k})$ , and the right-hand side is $ι (\sum_{k < σ (n)}^{N} (\binom{n}{k}))$ by the bridge clause of [L4], while the left-hand side is $ι (2)^{n} = ι (2^{n})$ by [L5]. Since $ι$ is injective, $\sum_{k < σ (n)}^{N} (\binom{n}{k}) = 2^{n}$ , which is step 1.1.

step 1.1L1L4L5

2.2

The hypothesis of clause 2 is not removable. At $n = 0$ the sum is $\sum_{k < 1} (- 1)^{k} ι (\binom{0}{k}) = (- 1)^{0} ι (\binom{0}{0}) = 1 \cdot 1 = 1 \neq 0$ , by [L6], $(\binom{0}{0}) = 1$ and $a^{0} = 1$ . What fails in the argument of step 1.2 is exactly one thing: $(- 1 + 1)^{0} = 0^{0} = 1$ rather than $0$ .

step 1.2L2L5L6

3.1

Clause 1 is step 1.1, confirmed by step 2.1; clause 2 is step 1.2, and step 2.2 shows why it carries the hypothesis $n \geq 1$ .

step 1.1step 1.2step 2.1step 2.2∎

Remarks

Two proofs of the same identity, deliberately. The counting proof is a statement about natural numbers and uses no embedding at all, while the analytic proof goes through $R$ and comes back by the injectivity of $ι$ . Recording both is what makes the agreement of the two readings visible rather than assumed.
Where the hypothesis of clause 2 is spent. In $0^{n} = 0$ , and nowhere else. The convention $0^{0} = 1$ is not a defect here: it is what makes the binomial theorem hold at $n = 0$ , and the price is that the alternating sum identity acquires a hypothesis. Both facts are consequences of the same convention.
The alternating sum is stated in $R$ because $(- 1)^{k}$ is not a natural number. The unsigned row sum, by contrast, is an identity between counts and is stated in $N$ .

Depends on

Used by

Dependency tree · two levels

61 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Binomial coefficient (Wikipedia) (standard reference, not scraped)
Binomial theorem (Wikipedia) (standard reference, not scraped)
Pascal's triangle (Wikipedia) (standard reference, not scraped)

∑k<n+1(nk)=2n∑k<n+1​(kn​)=2n, and ∑k<n+1(−1)kι ⁣(nk)=0∑k<n+1​(−1)kι(kn​)=0 for n≥1n≥1