Toolkit 14Sum of odd powern odd:an+bn=(a+b)(an−1−an−2b+⋯+bn−1)n\text{ odd}:\quad a^n+b^n=(a+b)\left(a^{n-1}-a^{n-2}b+\cdots+b^{n-1}\right)n odd:an+bn=(a+b)(an−1−an−2b+⋯+bn−1)ProofFor odd nnn, expanding the right-hand side gives(a+b)(an−1−an−2b+an−3b2−⋯−abn−2+bn−1)=(an−an−1b+an−2b2−⋯−a2bn−2+abn−1)+(an−1b−an−2b2+⋯+a2bn−2−abn−1+bn).\begin{aligned} &(a+b)\left(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \cdots - ab^{n-2} + b^{n-1}\right) \\ &= \left(a^n - a^{n-1}b + a^{n-2}b^2 - \cdots - a^2 b^{n-2} + ab^{n-1}\right) \\ &\quad + \left(a^{n-1}b - a^{n-2}b^2 + \cdots + a^2 b^{n-2} - ab^{n-1} + b^n\right). \end{aligned}(a+b)(an−1−an−2b+an−3b2−⋯−abn−2+bn−1)=(an−an−1b+an−2b2−⋯−a2bn−2+abn−1)+(an−1b−an−2b2+⋯+a2bn−2−abn−1+bn).All intermediate terms cancel, leavingan+bn.□a^n + b^n. \quad\squarean+bn.□Related ProblemsCoreAMC 10A/12A 2020 (Problem 21/19)AMC 10B 2022 (Problem 17)