Question

sn为数列{an}的前n项和,已知an>0,an^2+2an=4sn+3

Answer 1

(1)

根据an^2+2an=4Sn+3有：
a(n+1)^2+2a(n+1)=4S(n+1)+3
于是
an^2+2an = a(n+1)^2+2a(n+1)-4a(n+1)=a(n+1)^2-2a(n+1)
(an+1)^2 = [a(n+1)-1]^2

化简得到
a(n+1) = -an
a(n+1) = an +2

因为an>0,所以只有
a（n+1) = an+2 满足要求，也就是他是等差数列

又因为n=1时，a1^2 +2a1 = 4a1+3,a1 = 1

an = 1 + 2(n-1)=2n-1

(2)
bn = 1/(2n-1)(2n+1) = 0.5 *[1/(2n-1) -1/(2n+1)]

Sbn = b1 + b2 +....+bn
= 0.5(1/1-1/3) + 0.5(1/3-1/5) +....+0.5[1/(2n-1) -1/(2n+1)]
=0.5-0.5/(2n+1)

Answer 2

根据an^2+2an=4Sn+3有：
a(n+1)^2+2a(n+1)=4S(n+1)+3
于是
an^2+2an = a(n+1)^2+2a(n+1)-4a(n+1)=a(n+1)^2-2a(n+1)
(an+1)^2 = [a(n+1)-1]^2

化简得到
a(n+1) = -an
a(n+1) = an +2

因为an>0,所以只有
a（n+1) = an+2 满足要求，也就是他是等差数列

又因为n=1时，a1^2 +2a1 = 4a1+3,a1 = 1

an = 1 + 2(n-1)=2n-1

(2)
bn = 1/(2n-1)(2n+1) = 0.5 *[1/(2n-1) -1/(2n+1)]

Sbn = b1 + b2 +....+bn
= 0.5(1/1-1/3) + 0.5(1/3-1/5) +....+0.5[1/(2n-1) -1/(2n+1)]
=0.5-0.5/(2n+1)

Answer 3

根据an^2+2an=4Sn+3有：
a(n+1)^2+2a(n+1)=4S(n+1)+3
两式互减，可得a(n+1）^2-2a(n+1）=an^2+2an
两式两边加1，可得：[a(n+1)-1]^2=[an+1]^2
同时开方可得a(n+1)=an+2
将n=1带人题目中的式子可得a1=3
则an=3+2（n-1）=2n+1

2）
bn=1/ana(n+1)=1/(2n+1)(2n+3)=1/2[1/(2n+1)-1/(2n+3)]
则Sn=1/2[1/3-1/7+1/7-1/9……+1/(2n+1)-1/(2n+3)] (中间部分可抵消）
=1/2（1/3-1/2n+3）
=n/6n+9

Answer 4

Answer 5

(1)

根据an^2+2an=4Sn+3有：
a(n+1)^2+2a(n+1)=4S(n+1)+3
于是
an^2+2an = a(n+1)^2+2a(n+1)-4a(n+1)=a(n+1)^2-2a(n+1)
(an+1)^2 = [a(n+1)-1]^2

化简得到
a(n+1) = -an
a(n+1) = an +2

因为an>0,所以只有
a（n+1) = an+2 满足要求，也就是他是等差数列

又因为n=1时，a1^2 +2a1 = 4a1+3,a1 = 1

an = 1 + 2(n-1)=2n-1

(2)
bn = 1/(2n-1)(2n+1) = 0.5 *[1/(2n-1) -1/(2n+1)]

Sbn = b1 + b2 +....+bn
= 0.5(1/1-1/3) + 0.5(1/3-1/5) +....+0.5[1/(2n-1) -1/(2n+1)]
=0.5-0.5/(2n+1)