数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后面的是角标)

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数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后面的是角标)数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后

数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后面的是角标)
数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后面的是角标)

数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(2n+1),则an=?(字母后面的是角标)
由 a1+2a2+3a3+…+nan=n(n+1)(2n+1) 得
a1+2a2+3a3+…+nan+(n+1)a(n+1)=(n+1)(n+2)(2n+3)
两式相减得(n+1)a(n+1)=6(n+1)2
则a(n+1)=6(n+1) 故an=6n

a1+2a2+3a3+…+nan=n(n+1)(2n+1)
a1+2a2+3a3+…+(n-1)a(n-1)=(n-1)*n*(2n-1)
上述两式相减,有
nan=n(n+1)(2n+1)-n(n-1)(2n-1)
即 an=(2n^2+3n+1-2n^2+3n-1)=6n

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