Question

高三数学 求解前n项和

Answer 1

(1)
na(n+1)=n[S(n+1)-Sn]=Sn+n(n+1)
nS(n+1)-(n+1)Sn=n(n+1)
等式两边同除以n(n+1)
S(n+1)/(n+1)-Sn/n=1，为定值
S1/1=a1/1=2/1=2
数列{Sn/n}是以2为首项，1为公差的等差数列
Sn/n=2+1·(n-1)=n+1
Sn=n(n+1)
n≥2时，an=Sn-S(n-1)=n(n+1)-(n-1)n=2n
n=1时，a1=2·1=2，同样满足表达式
数列{an}的通项公式为an=2n
(2)
an/2?=2n/2?=n·???1
Tn=1·1+2·?+3·?2++n·???1
?Tn=1·?+2·?2++(n-1)·???1+n·??
Tn-?Tn=?Tn=1+?++???1-n·??
=1·(1-??)/(1-?) -n·??
=2-(n+2)·??
Tn=4-(n+2)·???1
(3)
bn=1/[ana(n+1)a(n+2)]
=1/[2n·2(n+1)·2(n+2)]
=(1/16)[1/n(n+1) -1/(n+1)(n+2)]
b1+b2++bn
=(1/16)[1/(1·2)-1/(2·3)+1/(2·3)-1/(3·4)++1/n(n+1)-1/(n+1)(n+2)]
=(1/16)[1/2 -1/(n+1)(n+2)]
=1/32 -1/[16(n+1)(n+2)]
1/[16(n+1)(n+2)]>0
1/32 -1/[16(n+1)(n+2)]<1/32
b1+b2++bn<1/32