Question

数学数列题求解

Answer 1

这样的答案

Answer 2

an = 2(Sn)^2/(2Sn -1) ( n>=2)
(I)
a1=1
for n>=2
an = Sn -S(n-1)
2(Sn)^2/(2Sn -1) = Sn -S(n-1)
2(Sn)^2 =2(Sn)^2 -2Sn.S(n-1) -Sn +S(n-1)
-2Sn.S(n-1) -Sn +S(n-1) =0
1/Sn - 1/S(n-1) = 2
=> {1/Sn} 是等差数列, d=2
1/Sn - 1/S1 = 2(n-1)
1/Sn - 1 = 2(n-1)
1/Sn = 2n-1
Sn = 1/(2n-1)
an = Sn - S(n-1)
=1/(2n-1) - 1/(2n-3)
ie
an
=1 ; n=1
=1/(2n-1) - 1/(2n-3) ; n=2,3,4,...
(II)

S1=a1 =1
for n>=2
Sn
=a1+(a2+a3+...+an)
=1+ { ( 1/3-1)+(1/5-1/3)+...+ [1/(2n-1) - 1/(2n-3)] }
= 1 + [ 1/(2n-1) - 1 ]
=1/(2n-1)
cn =Sn.S(n+1)
Tn = c1+c2+...+cn
T1=c1 = S1.S2 = 1 .( 1/3 ) = 1/3
for n>=2
cn
=Sn.S(n+1)
= 1/[(2n-1)(2n+1)]
= (1/2) [ 1/(2n-1) - 1/(2n+1) ]
Tn
=c1+c2+...+cn
=1/3 + (1/2) [ 1/3 - 1/(2n+1) ]
= 1/2 -1/[2(2n+1)]
=n/(2n+1) (1)
T1 = 1/3 满足 (1)
ie
Tn =n/(2n+1)