Sums of PowersLesson · BeginnerAlgebra: Sequences & Series1+2+3+⋯+n=n(n+1)2 1+2+3+\cdots+n=\frac{n(n+1)}{2} 1+2+3+⋯+n=2n(n+1)12+22+32+⋯+n2=n(n+1)(2n+1)6 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6} 12+22+32+⋯+n2=6n(n+1)(2n+1)13+23+33+⋯+n3=(n(n+1)2)2 1^3+2^3+3^3+\cdots+n^3= \left(\frac{n(n+1)}2\right)^2 13+23+33+⋯+n3=(2n(n+1))2Prerequisites (3)Gauss's Pairing Trick (Rainbow Idea)TelescopingAlgebraic IdentitiesSum of the first n integers1+2+3+⋯+n=n(n+1)2 1+2+3+\cdots+n=\frac{n(n+1)}2 1+2+3+⋯+n=2n(n+1)ProofBy Gauss Pairing TrickS=1+2+3+⋯+(n−2)+(n−1)+nS=n+(n−1)+(n−2)+⋯+3+2+12S=(n+1)+(n+1)+(n+1)+⋯+(n+1)+(n+1)+(n+1) \begin{array}{ccccccccccccccc} S&=&1&+&2&+&3&+&\cdots&+&(n-2)&+&(n-1)&+&n\\[8pt] S&=&n&+&(n-1)&+&(n-2)&+&\cdots&+&3&+&2&+&1\\[8pt] 2S&=&(n+1)&+&(n+1)&+&(n+1)&+&\cdots&+&(n+1)&+&(n+1)&+&(n+1) \end{array} 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=n(n+1)⇒S=n(n+1)2 \begin{aligned} \Rightarrow 2S&=n(n+1)\\[8pt] \Rightarrow S&=\frac{n(n+1)}2 \end{aligned} ⇒2S⇒S=n(n+1)=2n(n+1)Sum of squares12+22+⋯+n2=n(n+1)(2n+1)6 1^2+2^2+\cdots+n^2= \frac{n(n+1)(2n+1)}6 12+22+⋯+n2=6n(n+1)(2n+1)Proof(k+1)3−k3=3k2+3k+1(n+1)3−n3=3n2+3n+1n3−(n−1)3=3(n−1)2+3(n−1)+1⋮33−23=3(2)2+3(2)+123−13=3(1)2+3(1)+1 \begin{array}{ccccccccc} (k+1)^3&-&k^3&=&3k^2&+&3k&+&1\\[8pt] (n+1)^3&-&n^3&=&3n^2&+&3n&+&1\\[8pt] n^3&-&(n-1)^3&=&3(n-1)^2&+&3(n-1)&+&1\\[8pt] &&\vdots&&&&&&\\[8pt] 3^3&-&2^3&=&3(2)^2&+&3(2)&+&1\\[8pt] 2^3&-&1^3&=&3(1)^2&+&3(1)&+&1 \end{array} (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=3(12+22+⋯+n2)+3(1+2+⋯+n)+n⇒n3+3n2+3n=3(12+22+⋯+n2)+3n(n+1)2+n⇒12+22+⋯+n2=n3+3n2+3n3−n(n+1)2−n3=2n3+6n2+6n−3n2−3n−2n6=2n3+3n2+n6=n(2n2+3n+1)6=n(n+1)(2n+1)6 \begin{aligned} \Rightarrow\quad (n+1)^3-1^3 &= 3(1^2+2^2+\cdots+n^2) +3(1+2+\cdots+n)+n\\[8pt] \Rightarrow\quad n^3+3n^2+3n &= 3(1^2+2^2+\cdots+n^2) +\frac{3n(n+1)}2+n\\[8pt] \Rightarrow\quad 1^2+2^2+\cdots+n^2 &= \frac{n^3+3n^2+3n}{3} -\frac{n(n+1)}2-\frac n3\\[8pt] &= \frac{2n^3+6n^2+6n-3n^2-3n-2n}{6}\\[8pt] &= \frac{2n^3+3n^2+n}{6} = \frac{n(2n^2+3n+1)}6\\[8pt] &= \frac{n(n+1)(2n+1)}6 \end{aligned} ⇒(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)Sum of cubes13+23+⋯+n3=(n(n+1)2)2 1^3+2^3+\cdots+n^3= \left(\frac{n(n+1)}2\right)^2 13+23+⋯+n3=(2n(n+1))2Proof(k+1)4−k4=4k3+6k2+4k+1(n+1)4−n4=4n3+6n2+4n+1n4−(n−1)4=4(n−1)3+6(n−1)2+4(n−1)+1⋮34−24=4(2)3+6(2)2+4(2)+124−14=4(1)3+6(1)2+4(1)+1 \begin{array}{ccccccccccc} (k+1)^4&-&k^4&=&4k^3&+&6k^2&+&4k&+&1\\[8pt] (n+1)^4&-&n^4&=&4n^3&+&6n^2&+&4n&+&1\\[8pt] n^4&-&(n-1)^4&=&4(n-1)^3&+&6(n-1)^2&+&4(n-1)&+&1\\[8pt] &&\vdots&&&&&&&&\\[8pt] 3^4&-&2^4&=&4(2)^3&+&6(2)^2&+&4(2)&+&1\\[8pt] 2^4&-&1^4&=&4(1)^3&+&6(1)^2&+&4(1)&+&1 \end{array} (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=4(13+23+⋯+n3)+6(12+22+⋯+n2)+4(1+2+⋯+n)+n⇒n4+4n3+6n2+4n=4(13+23+⋯+n3)+n(n+1)(2n+1)+2n(n+1)+n⇒n4+4n3+6n2+4n=4(13+23+⋯+n3)+2n3+3n2+n+2n2+2n+n⇒4(13+23+⋯+n3)=n4+2n3+n2=n2(n2+2n+1)=n2(n+1)2 \begin{aligned} (n+1)^4-1^4 &= 4(1^3+2^3+\cdots+n^3) +6(1^2+2^2+\cdots+n^2) +4(1+2+\cdots+n)+n\\[8pt] \Rightarrow\quad n^4+4n^3+6n^2+4n &= 4(1^3+2^3+\cdots+n^3) +n(n+1)(2n+1) +2n(n+1)+n\\[8pt] \Rightarrow\quad n^4+4n^3+6n^2+4n &= 4(1^3+2^3+\cdots+n^3) +2n^3+3n^2+n +2n^2+2n+n\\[8pt] \Rightarrow\quad 4(1^3+2^3+\cdots+n^3) &= n^4+2n^3+n^2\\[8pt] &= n^2(n^2+2n+1)\\[8pt] &= n^2(n+1)^2 \end{aligned} (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)213+23+⋯+n3=(n(n+1)2)2 1^3+2^3+\cdots+n^3 = \left(\frac{n(n+1)}2\right)^2 13+23+⋯+n3=(2n(n+1))2Related Problems (6)AMC 10A 2020 (Problem 8)AMC 12A 2023 (Problem 12)AMC 10B 2025 (Problem 18)AMC 10B Fall 2021 (Problem 22)AMC 12A 2022 (Problem 16)AIME I 2025 (Problem 5)Next Toolkits (1)Arithmetic Sequences and Series