已知数列an满足a1+2a2+3a3+...+nan=n(n+1)*(n+2),则数列an的前n项和Sn=?
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已知数列an满足a1+2a2+3a3+...+nan=n(n+1)*(n+2),则数列an的前n项和Sn=?
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
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
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