13+ 23+ ⋯ + 𝑛3
12(2 − 1) + 22(3 − 1) + ⋯ + 𝑛2(𝑛 + 1 − 1)
12 2 − 12+ 22 3 − 22 + ⋯ + 𝑛2(𝑛 + 1) − 𝑛2
12 2 + 22 3 + ⋯ + 𝑛2(𝑛 + 1) − (12+ 22 + ⋯ + 𝑛2)
1.2(3 − 2) + 2.3(4 − 2) + ⋯ + 𝑛(𝑛 + 1)(𝑛 + 2 − 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) 4 = 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3) − 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 − 1)
𝑣ậ𝑦 đỏ 𝑡ℎà𝑛ℎ = 𝑛(𝑛 + 1)(𝑛 + 2)(𝑛 + 3)
4
12+ 22+ ⋯ + 𝑛2 = 1 (2 − 1) + 2 (3 − 1) + ⋯ + 𝑛(𝑛 + 1 − 1)
= 1.2 + 2.3 + ⋯ + 𝑛(𝑛 + 1) − (1 + 2 + ⋯ + 𝑛)(𝑡í𝑛ℎ 𝑐ũ𝑛𝑔 𝑡𝑡 𝑛ℎư 𝑡𝑟ê𝑛)
= 𝑛(𝑛+1)(2𝑛+1)
6
𝑘ế𝑡 ℎợ𝑝 𝑙ạ𝑖 𝑡𝑎 đượ𝑐:
𝑆 = 𝑛(𝑛+1)(𝑛+2)(𝑛+3)
4 −𝑛(𝑛+1)(2𝑛+1)
6 −2𝑛(𝑛+1)(𝑛+2)
3 =𝑛2(𝑛+1)2
4 =>
𝑆 𝑙à 𝑠ố 𝑐ℎí𝑛ℎ 𝑝ℎươ𝑛𝑔