bn = 2^n.an.a(n+1)
= 2^n/[(2^n-1)(2^(n+1) -1)]
= 1/(2^n-1) -1/(2^(n+1) -1)
Sn=b1+b2+...+bn
= 1/(2-1) - 1/(2^(n+1) -1)
= 1- 1/(2^(n+1) -1)
Sn < 1 好像是这样
已知an=1/(2^n-1)若数列bn满足bn=2^n an a(n+1),Sn=b1+b2+b3+……bn,求证Sn
已知an=1/(2^n-1)若数列bn满足bn=2^n an a(n+1),Sn=b1+b2+b3+……bn,求证Sn
a1=1
a1=1
数学人气:690 ℃时间:2020-05-16 13:51:32
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