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  "body": "\n## Statement\n\nLet $G$ be an additive abelian group, let $F \\colon \\Z \\to G$ be a function, and let $\\Delta F(k) = F(k + 1) - F(k)$ be its [[forward difference|TXTR68V4R1]].\nThen for all integers $a \\le b$,\n$$\\sum_{k=a}^{b-1} \\Delta F(k) = F(b) - F(a).$$\nEquivalently, if $F$ is an [[indefinite sum|5VK9NZM0R7]] of $f \\colon \\Z \\to G$, then $\\sum_{k=a}^{b-1} f(k) = F(b) - F(a)$.\n\n## Remarks\n\nThe proof is a telescoping sum: $\\sum_{k=a}^{b-1} (F(k+1) - F(k))$ cancels to $F(b) - F(a)$, as in a [[telescoping series|0GXKM1R7E4]].\nFor $a = b$ both sides are $0$.\nExample: $\\Delta (k)_3 = 3 (k)_2$ for the falling factorial $(k)_3 = k(k-1)(k-2)$, so $\\sum_{k=0}^{n-1} k(k-1) = \\frac{n(n-1)(n-2)}{3}$.\nA related identity, with $\\Delta$ of step $h$ on any additive commutative monoid, is $\\sum_{k=0}^{n-1} f(x + kh) = \\sum_{j=0}^{n-1} \\binom{n}{j+1} \\Delta_h^j f(x)$, obtained by summing [[Newton's forward difference formula|DDHCMKHD2T]].\n\n## Explanations\n\n### L1 human\n\nAdding up all the steps of a function gives its total change.\nThis is the discrete version of $\\int_a^b F'(x)\\,dx = F(b) - F(a)$.\nTo sum $f(a) + \\dots + f(b-1)$, it is enough to find one function $F$ whose steps are the values of $f$.\n\n### L2 human\n\nThe theorem reduces the evaluation of finite sums to finding indefinite sums, just as the [[fundamental theorem of calculus|0NHGY0N0JB]] reduces integrals to antiderivatives.\nA table of differences, such as $\\Delta (k)_m = m (k)_{m-1}$ for [[falling factorials|0SJNCJCCQG]] and $\\Delta c^k = (c - 1) c^k$, becomes a table of sums.\nTo sum $k^2$, write $k^2 = (k)_2 + (k)_1$, which gives $\\sum_{k=0}^{n-1} k^2 = \\frac{(n)_3}{3} + \\frac{(n)_2}{2}$.\n[[Summation by parts|9G82Z17DKA]] is the discrete analogue of integration by parts and is proved from the same telescoping.\n",
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