Question

利用差分性质证明1×3+2×4+…+n×(n+2)=[n(n+1)(2n+7)]/6

Answer 1

原式=(1^2＋2)＋(2^2＋4)＋...＋n^2＋2n=n(n＋1)(2n＋1)/6＋n(2n＋2)/2=n(n＋1)[(2n＋1)/6＋1]=n(n＋1)(2n＋7)/6.