数列解答题详细 5
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a1=1
(an- a(n+1))/an = 2a(n+1)
(1)
(an- a(n+1))/an = 2a(n+1)
an- a(n+1) = 2an.a(n+1)
1/(an+1) -1/an = 2
=>{1/an} 是等差数列, d=2
1/an -1/a1 =2(n-1)
1/an = 2n-1
an =1/(2n-1)
(2)
bn=an.a(n+1)
=1/[(2n-1)(2n+1)]
=(1/2) [ 1/(2n-1) -1/(2n+1)]
Tn=b1+b2+...+bn
=(1/2) [ 1/2 -1/(2n+1)]
= n/[2(2n+1)]
(an- a(n+1))/an = 2a(n+1)
(1)
(an- a(n+1))/an = 2a(n+1)
an- a(n+1) = 2an.a(n+1)
1/(an+1) -1/an = 2
=>{1/an} 是等差数列, d=2
1/an -1/a1 =2(n-1)
1/an = 2n-1
an =1/(2n-1)
(2)
bn=an.a(n+1)
=1/[(2n-1)(2n+1)]
=(1/2) [ 1/(2n-1) -1/(2n+1)]
Tn=b1+b2+...+bn
=(1/2) [ 1/2 -1/(2n+1)]
= n/[2(2n+1)]
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