Question [!problem] Prove by induction that, for n∈Z+n \in \mathbb{Z}^+n∈Z+ ∑r=1nr(2r−1)=16n(n+1)(3n+1)\sum_{r=1}^{n} r(2r-1) = \frac{1}{6}n(n+1)(3n+1)∑r=1nr(2r−1)=61n(n+1)(3n+1) (5) 中文翻译 用数学归纳法证明，对于 n∈Z+n \in \mathbb{Z}^+n∈Z+ ∑r=1nr(2r−1)=16n(n+1)(3n+1)\sum_{r=1}^{n} r(2r-1) = \frac{1}{6}n(n+1)(3n+1)∑r=1nr(2r−1)=61n(n+1)(3n+1) (5)
