Question

问一道数列题？

Answer 1

a1=1
a2=a1+q=1+q
a5=a1+4q=1+4q
a1，a2，a5成等比数列
∴a2:a1=a5:a2
(1+q)/1=(1+4q)/(1+q)

解方程得：q=2（公差不为0）
an=1+2a(n-1)
Sn=n(a1)+[n(n-1)/2]*q=n²

Answer 2

(1)
a1=1
an=a1+(n-1)d
a1.a5 = (a2)^2
(1+4d)=(1+d)^2
d^2-2d=0
d(d-2)=0
d=2
ie
an = 1+2(n-1) =2n-1
Sn = n^2
(2)
bn
=1/[ [a(n+1)]^2 -1]
=1/[ (2n+1)^2 -1]
=1/( 4n^2 +4n)
=(1/4)[ 1/n -1/(n+1)]
Tn
=b1+b2+...+bn
=(1/4) [ 1 - 1/(n+1)]
=n/[4(n+1)]