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Algebra Difficulty 5.0 AIME Prove it Taiwan

The definition of the sum of equal powers of consecutive positive integers is
Sm(N)=∑k=1N−1km S_m(N) = \sum_{k=1}^{N-1} k^m
In particular
S1(N)=1+2+⋯+(N−1)=N(N−1)2 S_1(N) = 1 + 2 + \cdots + (N-1) = \frac{N(N-1)}{2}
S2(N)=12+22+⋯+(N−1)2=N(N−1)(2N−1)6 S_2(N) = 1^2 + 2^2 + \cdots + (N-1)^2 = \frac{N(N-1)(2N-1)}{6}
S3(N)=13+23+⋯+(N−1)3=N2(N−1)24 S_3(N) = 1^3 + 2^3 + \cdots + (N-1)^3 = \frac{N^2(N-1)^2}{4}
Prove that
S2(N)S3(N)=712S6(N)+512S4(N) S_2(N)S_3(N) = \frac{7}{12}S_6(N) + \frac{5}{12}S_4(N)

Solution

S2(N)S3(N)=∑j=1N−1j2∑k=1N−1k3 S_2(N)S_3(N) = \sum_{j=1}^{N-1} j^2 \sum_{k=1}^{N-1} k^3
Split into three parts according to k<jk < j, k=jk = j, and k>jk > j
that is
∑j=1N−1j2∑k=1j−1k3+S5(N)+∑k=1N−1k2∑j=1k−1j2 \sum_{j=1}^{N-1} j^2 \sum_{k=1}^{j-1} k^3 + S_5(N) + \sum_{k=1}^{N-1} k^2 \sum_{j=1}^{k-1} j^2
or
∑j=1N−1j2⋅j2(j−1)24+S5(N)+∑k=1N−1k3⋅k(k−1)(2k−1)6 \sum_{j=1}^{N-1} j^2 \cdot \frac{j^2(j-1)^2}{4} + S_5(N) + \sum_{k=1}^{N-1} k^3 \cdot \frac{k(k-1)(2k-1)}{6}
the sum is
712S6(N)−S5(N)+512S4(N)+S5(N) \frac{7}{12}S_6(N) - S_5(N) + \frac{5}{12}S_4(N) + S_5(N)
that is
712S6(N)+512S4(N) \frac{7}{12}S_6(N) + \frac{5}{12}S_4(N)

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.