Question

已知数列｛an}前n项和Sn=n^2+n,令bn=1/an*a(n+1),求数列｛bn}的前n项和Tn

Answer 1

n>=2
an=Sn-S(n-1)=2n
a1=S1=2,也符合
所以an=2n
则bn=1/[2n(2n+2)]
=1/4*1/n(n+1)
=1/4*[(n+1)-n]/n(n+1)
=1/4[(n+1)/n(n+1)-n/n(n+1)]
=1/4[1/n-1/(n+1)]
所迹乎以Tn=1/4*[1-1/局旦2+1/2-1/3+……+1/n-1/姿腊悉(n+1)]
=1/4[1-1/(n+1)]
=n/(4n+4)

Answer 2

(3cos@-1)(sin@+3)=0
-1<灶差=sin@<=1
所以sin@=1/3
所袭辩棚以拍则cos2@=1-2(sin@)^2=7/9
3(x-2)2-12=0
(x-2)2=4
x-2=±2
x=2-2,x=2+2
x=0,x=4