Question

高中数列，求解

Answer 1

解：
(1)
2a2+a3+a5=20
2(a1+d)+a1+2d+a1+4d=20
4a1+8d=20
a1+2d=5 ①
S10=100
10a1+10·9d/2=100
a1+4.5d=10 ②
联立①、②，解得a1=1，d=2
an=a1+(n-1)d=1+2(n-1)=2n-1
数列{an}的通项公式为an=2n-1
(2)
bn=1/[ana(n+1)]=1/[(2n-1)(2n+1)]=½[1/(2n-1) -1/(2n+1)]
Tn=b1+b2+...+bn
=½[1/1 -1/3 +1/3 -1/5 +...+1/(2n-1) -1/(2n+1)]
=½[1- 1/(2n+1)]
=n/(2n+1)

Answer 2

2a2+a3+a5=2(a1+d)+(a1+2d)+(a1+4d)=20
4a1+8d=20
a1+2d=5 ........................................(1)
S10=10a1+[10(10-1)/2]d=100
a1+(9/2)d=10...................................(2)
(2)-(1)得：
(5d/2)=5
d=2
a1=1
an=2n-1
(2)
bn=1/(2n-1)(2n+1)=(1/2)[(1/2n-1)-(1/2n+1)]
Tn=(1/2)[(1-1/3)+(1/3)-(1/5)+(1/5)-(1/7)+.......+(1/2n-1)-(1/2n+1)]
=(1/2)[1-(1/2n+1)]
=n/(2n+1)