Question

高三数学数列题，求各位大神帮解•﹏•

Answer 1

(I)
a1=1/2
2an.a(n+1)= an - a(n+1)
1/a(n+1) -1/an = 2
=> {1/an } 是等差数列, d=2
1/an -1/a1 =2(n-1)
1/an =2n
an =1/(2n)

(II)
Sn =b1+b2+...+bn
bn= 1/[√(n-1) +√(n+1)] ; n=2k-1
=a(n/2). a(n/2 +1) ; n=2k

if n is odd
bn = 1/[√(n-1) +√(n+1)]
= (1/2) [√(n+1) - √(n-1)]

if n is even
bn = a(n/2). a(n/2 +1)
= (1/n).[ 1/(n+2) ]
=(1/2) [ 1/n - 1/(n+2) ]

if n is odd
Sn
=b1+b2+...+bn
=(b1+b3+...+bn) +(b2+b4+...+b(n-1))
=(1/2)√(n+1) + (1/2)[ 1/2 - 1/(n+1) ]

if n is even
Sn
=b1+b2+...+bn
=(b1+b3+...+b(n-1)) +(b2+b4+...+bn)
=(1/2)√n + (1/2)[ 1/2 - 1/(n+2) ]

(III)
Tn =1/a1+1/a2+...+1/an
= 2+4+...+(2n)
=n(n+1)

Tn/(n+c)
=n(n+1)/(n+c)
c=1 or 0
{Tn} 是等差数列

Answer 2

同除an an+1

Answer 3

大神来了

追答