作业帮sn为.{4^n-2^n}前n项和,bn=2^n/sn求{bn}前n项和Tn
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sn为.{4^n-2^n}前n项和,bn=2^n/sn求{bn}前n项和Tn
sn为.{4^n-2^n}前n项和,bn=2^n/sn求{bn}前n项和Tn
sn为.{4^n-2^n}前n项和,bn=2^n/sn求{bn}前n项和Tn
令：an=4^n,dn=2^n
则：cn=an-bn=4^n-2^n
Sn=(4+4^2+...+4^n)-(2+2^2+...+2^n)
an的前n项和：S1=4(1-4^n)/(1-4)=4(4^n-1)/3
dn的前n项和：S2=2(1-2^n)/(1-2)=2(2^n-1)
故：Sn=S1-S2=4*4^n/3-2^(n+1)+2/3
=4^(n+1)/3-2^(n+1)+2/3
=2^(2n+2)/3-2^(n+1)+2/3
故：bn=2^n/Sn=3*2^n/(2^(2n+2)-3*2^(n+1)+2)
=3*2^n/((2^(n+1)-1)(2^(n+1)-2))
=3/(2^(n+1)-2)-3/(2^(n+2)-2)
即bn的前一项的后面分项等于后一项的前面分项(不考虑符号)
故：Tn=3/2-3/(2^(n+2)-2)
=(3/2)(1-1/(2^(n+1)-1))
an=4^n-2^n
Sn=4^n+...+4-(2^n+...+2)=4(1-4^n)/(1-4)-2(1-2^n)/(1-2)=4(4^n-1)/3-2(2^n-1)
=4^(n+1)/3-2^(n+1)+2/3=[(2^(n+1))^2-3*2^(n+1)+2]/3
=[2^(n+1)-1][2^(n+1)-2]/3
bn=2^n/sn=3*2^n/[2^(n...
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an=4^n-2^n
Sn=4^n+...+4-(2^n+...+2)=4(1-4^n)/(1-4)-2(1-2^n)/(1-2)=4(4^n-1)/3-2(2^n-1)
=4^(n+1)/3-2^(n+1)+2/3=[(2^(n+1))^2-3*2^(n+1)+2]/3
=[2^(n+1)-1][2^(n+1)-2]/3
bn=2^n/sn=3*2^n/[2^(n+1)-1][2^(n+1)-2]=3*2^n*{1/[(2^(n+1)-2]-1/[2^(n+1)-1]}
=3{2^n/[(2^(n+1)-2]-*2^n/[2^(n+1)-1]}
b(n-1)=3{2^(n-1)/[(2^n-2]-*2^(n-1)/[2^n-1]}=3{2^n/[(2^(n+1)-4]-*2^n/[2^(n+1)-2]}
b(n-2)=3{2^n/[(2^(n+1)-2^(2+1)]-*2^n/[2^(n+1)-2^2]}
...
b1=b(n-(n-1))=3{2^n/[(2^(n+1)-2^n]-*2^n/[2^(n+1)-2^(n-1)]}
Tn=b1+b2+...+bn=3{2^n/[(2^(n+1)-2^n]-*2^n/[2^(n+1)-1]}
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