Question

已知等差数列{an}中,公差d>0,其前n项和为Sn,且满足:a2*a3=45,a1+a4=14

Answer 1

因为是等差数列，
所以，a1+a4=a3+a2=14
所以a3=14-a2，
将a3=14-a2代入a2a3=45得
（14-a2)*a2=45
a2^2-14a2+45=0
(a2-5)(a2-9)=0
a2=5或a2=9
a3=14-a2=14-5=9或a3=14-a2=14-9=5
d=a3-a2=9-5=4或d=a3-a2=5-9=-4(公差d>0,舍去)
所以，d=4
a1=a2-d=5-4=1
an=1+(n-1)*4=4n-3
(2)sn=n（a1+an）/2
Bn=Sn/（n+c）
Bn=n（a1+an）/[2（n+c）]
b1=1/(2+2c)
b2=2(1+5)/[2(2+c)]=6/(2+c)
b3=3(1+9)/[2(3+c)]=15/(3+c)
2b2=b1+b3
2*6/(2+c)=15/(3+c)+1/(2+2c)
12/(2+c)=15/(3+c)+1/(2+2c)

Answer 2

1、
略
2、
a(n+1)=4n+1
bn=1/4*4/(4n-3)(4n+1)
=1/4*[(4n+1)-(4n-3)]/[(4n-3)(4n+1)]
=1/4*{(4n+1)/[(4n-3)(4n+1)]-(4n-3)/[(4n-3)(4n+1)]}
=1/4*[1/(4n-3)-1/(4n+1)]
所以Sn=1/4*[1/1-1/5+1/5-1/9+1/9-1/13+……+1/(4n-3)-1/(4n+1)]
==1/4*[1-1/(4n+1)]
=n/(4n+1)