(* Title: HOL/ex/NatSum.thy Author: Tobias Nipkow *) section ‹Summing natural numbers› theory NatSum imports Main begin text ‹ Summing natural numbers, squares, cubes, etc. Thanks to Sloane's On-Line Encyclopedia of Integer Sequences, 🌐‹http://oeis.org›. › lemmas [simp] = ring_distribs diff_mult_distrib diff_mult_distrib2 ―‹for type nat› text ‹┉ The sum of the first ‹n› odd numbers equals ‹n› squared.› lemma sum_of_odds: "(∑i=0..<n. Suc (i + i)) = n * n" by (induct n) auto text ‹┉ The sum of the first ‹n› odd squares.› lemma sum_of_odd_squares: "3 * (∑i=0..<n. Suc(2*i) * Suc(2*i)) = n * (4 * n * n - 1)" by (induct n) auto text ‹┉ The sum of the first ‹n› odd cubes.› lemma sum_of_odd_cubes: "(∑i=0..<n. Suc (2*i) * Suc (2*i) * Suc (2*i)) = n * n * (2 * n * n - 1)" by (induct n) auto text ‹┉ The sum of the first ‹n› positive integers equals ‹n (n + 1) / 2›.› lemma sum_of_naturals: "2 * (∑i=0..n. i) = n * Suc n" by (induct n) auto lemma sum_of_squares: "6 * (∑i=0..n. i * i) = n * Suc n * Suc (2 * n)" by (induct n) auto lemma sum_of_cubes: "4 * (∑i=0..n. i * i * i) = n * n * Suc n * Suc n" by (induct n) auto text ‹┉ A cute identity:› lemma sum_squared: "(∑i=0..n. i)^2 = (∑i=0..n. i^3)" for n :: nat proof (induct n) case 0 show ?case by simp next case (Suc n) have "(∑i = 0..Suc n. i)^2 = (∑i = 0..n. i^3) + (2*(∑i = 0..n. i)*(n+1) + (n+1)^2)" (is "_ = ?A + ?B") using Suc by (simp add: eval_nat_numeral) also have "?B = (n+1)^3" using sum_of_naturals by (simp add: eval_nat_numeral) also have "?A + (n+1)^3 = (∑i=0..Suc n. i^3)" by simp finally show ?case . qed text ‹┉ Sum of fourth powers: three versions.› lemma sum_of_fourth_powers: "30 * (∑i=0..n. i * i * i * i) = n * Suc n * Suc (2 * n) * (3 * n * n + 3 * n - 1)" proof (induct n) case 0 show ?case by simp next case (Suc n) then show ?case by (cases n) ― ‹eliminates the subtraction› simp_all qed text ‹ Two alternative proofs, with a change of variables and much more subtraction, performed using the integers. › lemma int_sum_of_fourth_powers: "30 * int (∑i=0..<m. i * i * i * i) = int m * (int m - 1) * (int(2 * m) - 1) * (int(3 * m * m) - int(3 * m) - 1)" by (induct m) simp_all lemma of_nat_sum_of_fourth_powers: "30 * of_nat (∑i=0..<m. i * i * i * i) = of_nat m * (of_nat m - 1) * (of_nat (2 * m) - 1) * (of_nat (3 * m * m) - of_nat (3 * m) - (1::int))" by (induct m) simp_all text ‹┉ Sums of geometric series: ‹2›, ‹3› and the general case.› lemma sum_of_2_powers: "(∑i=0..<n. 2^i) = 2^n - (1::nat)" by (induct n) auto lemma sum_of_3_powers: "2 * (∑i=0..<n. 3^i) = 3^n - (1::nat)" by (induct n) auto lemma sum_of_powers: "0 < k ⟹ (k - 1) * (∑i=0..<n. k^i) = k^n - 1" for k :: nat by (induct n) auto end