已知数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2),则a1+a2+a3+…+an=多少?
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已知数列{an}满足a1+2a2+3a3+…+nan=n(n+1)(n+2),则a1+a2+a3+…+an=多少?
证明:
因为:a1+2a2+3a3+…+nan=n(n+1)(n+2),
记:bn=nan,那么:b1+b2+...+bn=n(n+1)(n+2)
将n-1带入,得:
b1+b2+...+b(n-1)=(n-1)n(n+1)
相减,得:
bn=n(n+1)(n+2)-(n-1)n(n+1)
=3n(n+1)
所以:nan=bn=3n(n+1)
所以:an=3n+3
所以:a1+a2+a3+…+an=(3n^2+9n)/2
因为:a1+2a2+3a3+…+nan=n(n+1)(n+2),
记:bn=nan,那么:b1+b2+...+bn=n(n+1)(n+2)
将n-1带入,得:
b1+b2+...+b(n-1)=(n-1)n(n+1)
相减,得:
bn=n(n+1)(n+2)-(n-1)n(n+1)
=3n(n+1)
所以:nan=bn=3n(n+1)
所以:an=3n+3
所以:a1+a2+a3+…+an=(3n^2+9n)/2
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