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共回答了17个问题采纳率:100% 举报
(1)
an=aⁿ
bn=anlgan=aⁿlg(aⁿ)=naⁿlga
Sn=b1+b2+...+bn
=1×a×lga+2×a²lga+...+naⁿlga
=(a+2a²+...+naⁿ)lga
令Cn=a+2a²+...+naⁿ
则Cn×a=a²+2a³+...+(n-1)aⁿ+na^(n+1)
Cn-Cn×a=a+a²+a³+...+aⁿ-na^(n+1)=a(aⁿ-1)/(a-1)-na^(n+1)
(1-a)Cn=a(aⁿ-1)/(a-1)-na^(n+1)
Cn=na^(n+1)/(a-1)-a(1-aⁿ)/(a-1)²
Sn=na^(n+1)lga/(a-1)-a(1-aⁿ)lga/(a-1)²
(2)
Sn/bn=[na^(n+1)lga/(a-1)-a(1-aⁿ)lga/(a-1)²]/(naⁿlga)
=a/(a-1)+a/[n(a-1)²]-1/[na^(n-1)(a-1)²]
1年前
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