$\sum_{j<m+1}(-1)^{j}\,\iota\binom{t}{j} = (-1)^{m}\,\iota\binom{t-1}{m}$ for every $t \ge 1$ and every $m$

Let $t \in \mathbb{N}$ with $t \ge 1$ and let $m \in \mathbb{N}$. Then, in $\mathbb{R}$,

$\sum_{j<m+1}(-1)^{j}\,\iota\binom{t}{j} \;=\; (-1)^{m}\,\iota\binom{t-1}{m},$

where $\iota$ is the canonical natural (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), the binomial coefficients are the counts of The set $[A]^{k}$ of $k$-element subsets and the binomial coefficient $\binom{n}{k} := \lvert [n]^{k}\rvert$, and $t-1$ is the truncated difference of Finite sums and finite products of natural numbers, $\sum_{k<n} a_k$ and $\prod_{k<n} a_k$ in $\mathbb{N}$, which for $t \ge 1$ is the ordinary one, so that $(t-1) + 1 = t$.

The hypothesis $t \ge 1$ is part of the statement. At $t = 0$ and $m = 1$ the left-hand side is $\iota\binom{0}{0} - \iota\binom{0}{1} = 1$, while the truncated difference gives $t - 1 = 0$ and the right-hand side is $-\iota\binom{0}{1} = 0$.

Two readings worth recording. At $m = 0$ both sides are $1$, since $\binom{t}{0} = \binom{t-1}{0} = 1$. For $m \ge t$ both sides are $0$: the terms of the left-hand side with $j > t$ vanish and the remaining sum is the full alternating row sum of $t$, which vanishes because $t \ge 1$, while $\binom{t-1}{m} = 0$ because $m > t-1$.

• Where $t \ge 1$ is spent. In exactly one place: the identity $s+1 = t$, which is what lets Pascal's rule be applied with upper index $s+1$. Under the truncated difference the equation $s+1 = t$ fails at $t = 0$, where $s = 0$ and $s+1 = 1$, and the statement fails there too.

• Why not the full alternating row sum. The published corollary of the binomial theorem gives the sum over the whole row, and only for $t \ge 1$. A truncation of that row is a different quantity, and the identity above is what says how far a truncation misses: by exactly one binomial coefficient of the row above, with the sign of the last term kept.