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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$sn为.{4^n-2^n}前n项和,bn=2^n/sn求{bn}前n项和Tn$

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))

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