设数列an通项an=4^n-1,数列bn通项bn=3n-1
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发布时间：2024-10-24 07:23
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时间：2024-10-25 14:30
let
S = 1.3^0+2.3^1+...+n.3^(n-1) (1)
3S = 1.3^1+2.3^2+...+n.3^n (2)
(2)-(1)
2S=n.3^n - ( 1+3+3^2+...+3^(n-1))
=n.3^n - (1/3)(3^n-1)
3S =(3/2)[n.3^n - (1/3)(3^n-1)]
cn=an.bn
= (3n-1).4^(n-1)
= 3(n.4^(n-1)) - 4^(n-1)
Sn =c1+c2+...+cn
=3S - (1/3)(4^n-1)
=(3/2)n.3^n - (5/2)(1/3)(3^n-1)
=5/6 + [(3/2)n - 5/6].3^n
