Question

数列{an}满足a1+2a2+3a3+.....+nan=n(n+1)(n+2),求数列{an}的前n项和Sn

Answer 1

a1+2a2+3a3+...+nan=n(n+1)*(n+2)，则：
a1+2a2+3a3+...+（n-1）×an-1=n(n-1)*(n+1),两式相减：
nan=n(n+1)*(n+2)-n(n-1)*(n+1)，得
an=3n+3
所以：
a1+a2+a3+...+an=3*(1+2+3+...+n)+3n=3*n(n+1)/2+3n
整理得前n项和为:a1+a2+a3+...+an=3n(n+3)/2

Answer 2

数列{an}满足a1+2a2+3a3+.....+nan=n(n+1)(n+2),求数列{an}的前n项和Sn

因为a1+2a2+3a3+…+nan=n×(n+1)×(n+2) ①
所以a1+2a2+3a3+…+(n-1)a(n-1)=(n-1)×n×(n+1) ②
①-②得
nan=n×(n+1)×(n+2)-(n-1)×n×(n+1) (n∈N*)
∴通项公式an=3n+3

前n项和Sn
Sn=a1+a2+a3+……+an
Sn=(3+3)+(3×2+3)+(3×3+3)+……+(3×n+3)
Sn=3×(1+2+3+……+n)+3n
Sn=3n(n+1)/2+3n
Sn=[3n(n+3)]/2

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Answer 3

a1=6
(1)a1+2a2+3a3+.....+nan=n(n+1)(n+2)
(2)a1+2a2+3a3+.....+(n-1)a(n-1)=(n-1)n(n+1)
(1)-(2)得
nan=n(n+1)(n+2)-(n-1)n(n+1)=3n(n+1)
an=3(n+1)
an-a(n-1)=3(n+1)-3n=3 d=3
sn=na1+n(n-1)/2*d=6n+3/2n(n-1)=3/2n^2+9/2n

Answer 4

a1+2a2+3a3+.....+nan=n(n+1)(n+2),①
以n-1代n,得a1+2a2+3a3+.....+(n-1)a<n-1>=(n-1)n(n+1),②
①-②，nan=3n(n+1),
∴an=3(n+1),
a1=6,
{an}是等差数列，
∴Sn=n[6+3(n+1)]/2=n(3n+9)/2.

Answer 5

(3/2)(n+3)