若数列{an}满足a1+2a2+3a3+~~+nan=n(n+1)(2n+1),则an=

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若数列{an}满足a1+2a2+3a3+~~+nan=n(n+1)(2n+1),则an=
若数列{an}满足a1+2a2+3a3+~~+nan=n(n+1)(2n+1),则an=

若数列{an}满足a1+2a2+3a3+~~+nan=n(n+1)(2n+1),则an=
a1+2a2+3a3+~+nan=n(n+1)(2n+1)知,
a1+2a2+3a3+~+(n-1)an-1=(n-1)n(2n-1),
n≠1时两式相减知an=(n+1)(2n+1)-(n-1)(2n-1)=6n,
n=1时a1=6,满足an=6n
所以an=6n

a1+2a2+3a3+~~+(n-1)a(n-1)+nan=n(n+1)(2n+1),
a1+2a2+3a3+~~+(n-1)a(n-1)=(n-1)(n-1+1)(2n-2+1)=（n-1)n(2n-1),(n>1)
相减 nan=n(n+1)(2n+1)-(n-1)n(2n-1)=n(2n^2+3n+1-2n^2+3n-1)=n*6n,
an=6n(n>1)
n=1,a1=1*2*3=6 符合
所以an=6n

a1+2a2+3a3+........+nan=n(n+1)(2n+1)
a1+2a2+3a3+........+nan+(n+1)an+1=(n+1)(n+2)(2n+3)
(n+1)an+1=(n+1)(6n+6)
an=6n

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a1+2a2+3a3+~~+nan=n(n+1)(2n+1), 得 sn +（a2 + 2a3 +......+(n-1)an）=n(n+1)(2n+1)
a1+2a2+3a3+~~+(n-1)a(n-1)=(n-1)n(2n-1), 得 s(n-1) +（a2 + 2a3 +......+(n-2)a(n-1)）=(n-1)n(2n-1),
所以 sn - s(n-1) +(n - 1)an =n(n+1)(2n+1) - (n-1)n(2n-1)
an+(n - 1)an =n(n+1)(2n+1) - (n-1)n(2n-1)
nan =n(n+1)(2n+1) - (n-1)n(2n-1)
an=(n+1)(2n+1) - (n-1)(2n-1)

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