数列1×4,2×5,3×6,…,n×(n+3),…则它的前n项和Sn=______.
∵an=n×(n+3)=n2+3n,
∴Sn=a1+a2+a3+…+an
=(1+3×1)+(4+3×2)+(9+3×3)+…+(n2+3n)
=(12+22+32+…+n2)+3(1+2+3+…+n)
=
+n(n+1)(2n+1) 6 3n(n+1) 2
=
.n(n+1)(n+5) 3
答案:
.n(n+1)(n+5) 3
数列1×4,2×5,3×6,…,n×(n+3),…则它的前n项和Sn=______.
∵an=n×(n+3)=n2+3n,
∴Sn=a1+a2+a3+…+an
=(1+3×1)+(4+3×2)+(9+3×3)+…+(n2+3n)
=(12+22+32+…+n2)+3(1+2+3+…+n)
=
+n(n+1)(2n+1) 6 3n(n+1) 2
=
.n(n+1)(n+5) 3
答案:
.n(n+1)(n+5) 3