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已知数列{an}的前n项和Sn=-an-(1/2)的n-1次方再加2,令bn=2的n次方乘an,求证bn等差,
还要求an
人气:218 ℃ 时间:2019-10-18 02:47:18
解答
Sn = -an + 2 - 2^(1-n)
S(n+1) = -a(n+1) + 2 - 2^(-n)
a(n+1)= -a(n+1) + an - 2^(-n) + 2^(1-n)
2a(n+1) = an + 2^(-n)
两边同乘以2的n次方
得到2^(n+1)·a(n+1) - 2^n·an = b(n+1) - bn = 1
S1 = a1 = -a1 +1 得 a1 = 0.5
b1 = 2a1 =1
bn = n = 2^n·an 得 an = 2^(-n)an=n/(2^n)呃 是我疏忽了不好意思
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