求数列{n(n+1)(n+2)}的前n项和sn

Question

在线等，急急！！

Answer 1

设数列{n(n+1)(n+2)(n+3)}的前n项和为Tn
Tn=1×2×3×4+2×3×4×5+3×4×5×6+…+n(n+1)(n+2)(n+3)
Tn= 1×2×3×4+2×3×4×5+…+(n-1)n(n+1)(n+2)+n(n+1)(n+2)(n+3)
相减得0=1×2×3×4+4[2×3×4+3×4×5+n(n+1)(n+2)]-n(n+1)(n+2)(n+3)
即0=24+4[Sn-1×2×3]-n(n+1)(n+2)(n+3)
解得Sn=n(n+1)(n+2)(n+3)/4

Answer 2

an=n^3+3n^2+2n
其实就是要知道n^3和n^2的求和公式
∑n^3=(1+2+...n)^2=[n(n+1)/2]^2=n^2(n+1)^2/4
∑n^2=n(n+1)(2n+1)/6
Sn=n^2(n+1)^2/4+n(n+1)(2n+1)/2+n(n+1)
=n(n+1)[n(n+1)/4+(2n+1)/2+1]
=n(n+1)(n^2+5n+6)/4
=n(n+1)(n+2)(n+3)/4