第11题，数列求和问题
1个回答
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(1)
2√Sn =an+1
4Sn = (an +1)^2
n=1
4a1 = (a1+1)^2
a1=1
an = Sn-S(n-1)
4an =(an +1)^2 - (a(n-1) +1)^2
(an)^2- (a(n-1))^2 - 2[an+ a(n-1)]=0
[an + a(n-1)].[an - a(n-1)-2]=0
an - a(n-1)-2=0
an - a1= 2(n-1)
an = 2n-1
(2)
bn = 1/[an.a(n+1)]
= (1/2)[ 1/(2n-1) - 1/(2n+1)]
Bn=b1+b2+...+bn
= (1/2) [ 1 - 1/(2n+1)]
=n/(2n+1)
2√Sn =an+1
4Sn = (an +1)^2
n=1
4a1 = (a1+1)^2
a1=1
an = Sn-S(n-1)
4an =(an +1)^2 - (a(n-1) +1)^2
(an)^2- (a(n-1))^2 - 2[an+ a(n-1)]=0
[an + a(n-1)].[an - a(n-1)-2]=0
an - a(n-1)-2=0
an - a1= 2(n-1)
an = 2n-1
(2)
bn = 1/[an.a(n+1)]
= (1/2)[ 1/(2n-1) - 1/(2n+1)]
Bn=b1+b2+...+bn
= (1/2) [ 1 - 1/(2n+1)]
=n/(2n+1)
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