Question

等差数列{a n }的各项均为正数,其前n项和为S n ,满足2S 2 =a 2 (a 2 +1),且a 1 =1.

Answer 1

(1) a _n=n (2) b ₄=

(1)设数列{a _n}的公差为d.

由2S ₂=

+a ₂,

可得2(a ₁+a ₁+d)=(a ₁+d) ²+(a ₁+d).

又a ₁=1,可得d=1(d=-2舍去),∴a _n=n.

(2)根据(1)得S _n=

,

b _n=

=

=n+

+1.

由于函数f(x)=x+

(x>0)在(0,

]上单调递减,在[

,+∞)上单调递增,

而3<

<4,且f(3)=3+

=

=

,

f(4)=4+

=

=

,

所以当n=4时,b _n取得最小值,

且最小值为

+1=

,

即数列{b _n}的最小值项是b ₄=

.