求数列1×4+1,2×5+1,3×6+1,……,n(n+3)+1的前n项和

Question

求数列 1×4+1,2×5+1,3×6+1,……,n(n+3)+1的前n项和

Answer 1

n(n+3)+1=n^2+3n+1
所以 1×4+1,2×5+1,3×6+1,……,n(n+3)+1的前n项和
=1^2+2^2+3^2+...+n^2+3(1+2+3+...+n)+n
=1/6n(n+1)(2n+1)+3n(n+1)/2+n
=1/6n(n+1)(2n+10)+n
=1/3n(n^2+6n+5+3)
=1/3n(n+2)(n+4)