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

来源：学生作业帮助网编辑：作业帮时间：2024/04/28 01:46:15

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
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

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
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)

周期性数列问题i已知数列｛an}满足a(n+1)=2an (0 已知数列an满足1/a-an=2根号n,且an>0.求an的通项公式是数列{an}满足1/an-an=2根号n，且an>0，求an的通项公式。已知数列an满足条件a1=-2 a（n+1）=2an/(1-an) 则an= 已知数列an满足a1=1,a(n+1)=an/(3an+2),则an=? 已知数列an满足a1=2,an=a(n-1)+2n,（n≥2）,求an 已知数列an满足:a1=1,an-a(n-1)=n n大于等于2 求an 已知数列{an}满足a1=33,a(n+1)-an=2n,则an/n的最小值已知数列{an}满足a1=33,a（n+1）-an=2n,求an/n的最小值已知数列an满足a1=100,a(n+1)-an=2n,则(an)/n的最小值为已知数列{an}满足a(n+1)=an+n,a1=1,则an= 已知数列{an}满足an+1=2an+3.5^n,a1=6.求an 已知数列{an}满足a1=2,a(n+1)-an=a(n+1）*an,则a31=? 已知数列{An}满足A(n+1)=【2An (0 已知数列｛an｝满足a(n+1)=an+3n+2,且a1=2,求an=? 已知数列an满足a(n+1)=an+3n+2,且a1=2,求an 已知数列｛An｝满足a1=1,a（n+1）=2an+1 求证数列{an+1}是等比数列求数列｛an｝通式已知数列{an}满足a(n+1)=an+lg2,a1=1,求an 已知数列｛an｝满足：a1=1,且an-a（n-1）=2n.求a2,a3,a4.求数列｛an｝通项an