1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 = n 2 + n 2 {\displaystyle 1+2+3+\cdots +n={n(n+1) \over 2}={n^{2}+n \over 2}} (số tam giác) 1 2 + 2 2 + 3 2 + ⋯ + n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 = 2 n 3 + 3 n 2 + n 6 {\displaystyle 1^{2}+2^{2}+3^{2}+\cdots +n^{2}={n(n+1)(2n+1) \over 6}={2n^{3}+3n^{2}+n \over 6}} (số hình chóp vuông (tiếng Anh là square pyramidal number)) 1 3 + 2 3 + 3 3 + ⋯ + n 3 = ( n 2 + n 2 ) 2 = n 4 + 2 n 3 + n 2 4 {\displaystyle 1^{3}+2^{3}+3^{3}+\cdots +n^{3}=\left({n^{2}+n \over 2}\right)^{2}={n^{4}+2n^{3}+n^{2} \over 4}} (số tam giác vuông (tiếng Anh là squared triangular number)) 1 4 + 2 4 + 3 4 + ⋯ + n 4 = 6 n 5 + 15 n 4 + 10 n 3 − n 30 {\displaystyle 1^{4}+2^{4}+3^{4}+\cdots +n^{4}={6n^{5}+15n^{4}+10n^{3}-n \over 30}} 1 5 + 2 5 + 3 5 + ⋯ + n 5 = 2 n 6 + 6 n 5 + 5 n 4 − n 2 12 {\displaystyle 1^{5}+2^{5}+3^{5}+\cdots +n^{5}={2n^{6}+6n^{5}+5n^{4}-n^{2} \over 12}} 1 6 + 2 6 + 3 6 + ⋯ + n 6 = 6 n 7 + 21 n 6 + 21 n 5 − 7 n 3 + n 42 {\displaystyle 1^{6}+2^{6}+3^{6}+\cdots +n^{6}={6n^{7}+21n^{6}+21n^{5}-7n^{3}+n \over 42}}