an=2∧n,bn=2n-1,设cn=an×bn,求数列cn的前n项和Tn
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let
S = 1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S=n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n -1)
cn = an.bn
=(2n-1)2^n
= 2.(n.2^n) -2^n
Tn = c1+c2+...+cn
=2S - 2(2^n -1)
=2n.2^(n+1) -6(2^n -1)
=6 +(4n-6).2^n
S = 1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S=n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n -1)
cn = an.bn
=(2n-1)2^n
= 2.(n.2^n) -2^n
Tn = c1+c2+...+cn
=2S - 2(2^n -1)
=2n.2^(n+1) -6(2^n -1)
=6 +(4n-6).2^n
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