Sums of Powers
Lesson · Beginner
Algebra: Sequences & Series
1+2+3+⋯+n=2n(n+1)
12+22+32+⋯+n2=6n(n+1)(2n+1)
13+23+33+⋯+n3=(2n(n+1))2
1+2+3+⋯+n=2n(n+1)
By Gauss Pairing Trick
SS2S===1n(n+1)+++2(n−1)(n+1)+++3(n−2)(n+1)+++⋯⋯⋯+++(n−2)3(n+1)+++(n−1)2(n+1)+++n1(n+1)
⇒2S⇒S=n(n+1)=2n(n+1)
12+22+⋯+n2=6n(n+1)(2n+1)
(k+1)3(n+1)3n33323−−−−−k3n3(n−1)3⋮2313=====3k23n23(n−1)23(2)23(1)2+++++3k3n3(n−1)3(2)3(1)+++++11111
⇒(n+1)3−13⇒n3+3n2+3n⇒12+22+⋯+n2=3(12+22+⋯+n2)+3(1+2+⋯+n)+n=3(12+22+⋯+n2)+23n(n+1)+n=3n3+3n2+3n−2n(n+1)−3n=62n3+6n2+6n−3n2−3n−2n=62n3+3n2+n=6n(2n2+3n+1)=6n(n+1)(2n+1)
13+23+⋯+n3=(2n(n+1))2
(k+1)4(n+1)4n43424−−−−−k4n4(n−1)4⋮2414=====4k34n34(n−1)34(2)34(1)3+++++6k26n26(n−1)26(2)26(1)2+++++4k4n4(n−1)4(2)4(1)+++++11111
(n+1)4−14⇒n4+4n3+6n2+4n⇒n4+4n3+6n2+4n⇒4(13+23+⋯+n3)=4(13+23+⋯+n3)+6(12+22+⋯+n2)+4(1+2+⋯+n)+n=4(13+23+⋯+n3)+n(n+1)(2n+1)+2n(n+1)+n=4(13+23+⋯+n3)+2n3+3n2+n+2n2+2n+n=n4+2n3+n2=n2(n2+2n+1)=n2(n+1)2
13+23+⋯+n3=(2n(n+1))2