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已知数列an满足a(n+1)=-an^2+2an,且a1=9/10,令bn=lg（1-an),求1/bn所有项和
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答案和解析
a = -an^2 + 2an
a - 1 = - an^2 + 2an - 1
a - 1 = - (an -1)^2
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n) = -an^2 + 2an
a - 1 = - an^2 + 2an - 1
a - 1 = - (an -1)^2
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n) - 1 = - an^2 + 2an - 1
a - 1 = - (an -1)^2
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n) - 1 = - (an -1)^2
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n)
a
a
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n)
a
a
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n)
a
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n)
a2 - 1 = - (a1 - 1)^2
a3 - 1 = - (a2 - 1)^2 = -(a1 -1)^4
a4 - 1 = - (a3 - 1)^2 = -(a1 - 1)^8
a5 - 1 = - (a4 - 1)^2 = -(a1 - 1)^16
余此类推,得到
an -1 = - (a1 - 1)^[2^(n-1)]
代入 a1 = 9/10
an - 1 = - (1/10)^[2^(n-1)]
1 - an = (1/10)^[2^(n-1)] = 10^[-2^(n-1)]
bn = lg(1 - an) = -2^(n-1)
1/bn = -1/2^(n-1)
1/b1 = -1
Sn = -[1 + 1/2 + 1/2^2 + …… + 1/2^(n-1)]
= - [1 - (1/2)^n]/(1 - 1/2)
= -2(1 - 1/2^n)
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