Let $$\sigma_k (n)= \sum_{j = 1}^{n} j^k$$ and $$\Psi_k (n)= \sum_{j = 1}^{n} \sigma_k (j).$$
Prove the following without use of the respective polynomial equations associated with $\sigma_k$ and $\Psi_k$. $$\sigma_1 (n) * \Psi_1 (n) = n \Psi_2 (n) .$$ $$$$ CONTEXT: When proving the famous theorem $\sigma_3 (n)= \sigma_1 ^2 (n)$ you may uncover the following identity $$\sigma_1 (n-1) + \sigma_1 (n) = n^2 .$$ Therefore (with $\sigma_k (0) = 0$ for all $k$) $$\sigma_2 (n) = \sum_{j=1}^{n} \bigg[\sigma_1 (j-1) + \sigma_1 (j)\bigg] .$$ Expanding the sum gives the result $$\sigma_2 (n) = \sigma_1 (n) + \sum_{j = 1}^{n-1} 2\sigma_1 (j)$$ which is where our interest in $\Psi_k$ begins. Using difference tables and linear algebra, it is relatively easy to find polynomial expressions for $\Psi_k$. I have found $$\Psi_1 (n) = \frac{n(n+1)(n+2)}{6}$$ $$\Psi_2 (n)= \frac{1}{12} n^{4} + \frac{1}{3} n^{3} + \frac{5}{12} n^{2} + \frac{1}{6} n .$$ I have found polynomial expressions for up to $\Psi_{13}$, but those are irrelevant in the context of this problem. I noticed that $\sigma_1 (n) * \Psi_1 (n) = n \Psi_2 (n)$ while fooling around with these and that is the motivation for the problem I have posted today. I'm not great with summations, but many users on this site have amazing insight into these kinds of problems, and was looking for some interesting takes. Thanks.