Question

高中数学 已知等差数列an的前n项为sn a1＝5 3a5＋a9＝s6

Answer 1

(1).s6=a1+...+a6=3a5+a9
又因为a1=5，a2=5+d,...,a6=5+5d
所以s6=6a1+15d=30+15d=3a5+a9=20+20d
d=2
an=5+2（n-1）=2n+3
(2).b1=a6=15
b(n+1)=a(n+1)an=(an+d)an
an=a1+2(n-1)
bn=an(an-d)=(a1+2n-2)(a1+2n-4)=25+10n-20+10n+4n^2-8n-10-4n+8=4n^2+8n+3=(2n+2)^2-1=(2n+3)(2n+1)
2/bn=1/(2n+1)-1/(2n+3)
所以1/bn的前n项和为
Tn=1/2*[1/3-1/5+1/5-1/7+1/7-1/9....+1/(2n+1)-1/(2n+3)]
=1/2*[1/3-1/(2n+3)]
=n/(6n+9)

Answer 2

3(a1+4d)+a1+8d=6a1+6*(6-1)*d/2
4a1+20d=6a1+15d
5d=2a1=10
d=2
an=a1+(n-1)d=2n+3