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数列an=2n-1,bn=2^(n-1)+1(n大于等于2),设Tn=a1/(b1-1)+a2/(b2-1)+……+an/(bn-1),求Tn
bangmanga
人气:372 ℃ 时间:2020-04-22 21:44:18
解答
(Bn)-1=2^(n-1)
Tn=1/1+3/2+5/4+……+(2n-1)/2^(n-1)
两边乘2
2Tn=2+3/1+5/2+……+(2n-1)/2^(n-2)
错位相减
Tn=2+2[1/1+1/2+1/4+……+1/2^(n-2)]-(2n-1)/2^(n-1)
=2+2(1-(1/2)^(n-1))/(1-1/2)-(2n-1)/2^(n-1)
=6-(2n+3)/2^(n-1)
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