số chính phương lớp 9

13 + 23 + ⋯ + 𝑛 12 (2 − 1) + 22 (3 − 1) + ⋯ + 𝑛2 (𝑛 + − 1) 12 − 12 + 22 − 22 + ⋯ + 𝑛2 (𝑛 + 1) − 𝑛2 12 + 22 + ⋯ + 𝑛2 (𝑛 + 1) − (12 + 22 + ⋯ + 𝑛2 ) 1.2(3 − 2) + 2.3(4 − 2) + ⋯ + 𝑛(𝑛 + 1)(𝑛 + − 2) − (12 + 22 + ⋯ + 𝑛2 ) 1.2.3 + 2.3.4 + ⋯ + 𝑛(𝑛 + 1)(𝑛 + 2) − 2(1.2 + 2.3 + ⋯ + 𝑛(𝑛 + 1)) − (12 + 22 + ⋯ + 𝑛2 ) 𝑡í𝑛ℎ 𝑡ℎằ𝑛𝑔 đỏ: 1.2.3.4 = 4.3.2.1 − 0.1.2.3 2.3.4.4 = 5.4.3.2 − 1.2.3.4 𝑛(𝑛 + 1)(𝑛 + 2) = 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3) − 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 − 1) 𝑣ậ𝑦 đỏ 𝑡ℎà𝑛ℎ = 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3) 12 + 22 + ⋯ + 𝑛2 = (2 − 1) + (3 − 1) + ⋯ + 𝑛(𝑛 + − 1) = 1.2 + 2.3 + ⋯ + 𝑛(𝑛 + 1) − (1 + + ⋯ + 𝑛)(𝑡í𝑛ℎ 𝑐ũ𝑛𝑔 𝑡𝑡 𝑛ℎư 𝑡𝑟ê𝑛) = 𝑛(𝑛+1)(2𝑛+1) 𝑘ế𝑡 ℎợ𝑝 𝑙ạ𝑖 𝑡𝑎 đượ𝑐: 𝑆= 𝑛(𝑛+1)(𝑛+2)(𝑛+3) − 𝑆 𝑙à 𝑠ố 𝑐ℎí𝑛ℎ 𝑝ℎươ𝑛𝑔 𝑛(𝑛+1)(2𝑛+1) − 2𝑛(𝑛+1)(𝑛+2) = 𝑛2 (𝑛+1)2 =>