已知数列an满足an>0,Sn=[(an+1)/2]^2,bn=(-1)^n*Sn,求b1+b2+……+bn

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4Sn=(an+1)^2
4S(n-1)=[a(n-1)+1]^2
an=Sn-S(n-1)
所以相减
4an=(an+1)^2-[a(n-1)+1]^2
(an+1)^2-4an=[a(n-1)+1]^2
(an-1)^2=[a(n-1)+1]^2
an-1=a(n-1)+1或an-1=-a(n-1)-1
an=a(n-1)+2或an=-a(n-1)
an>0
所以an=-a(n-1)不成立
所以an=a(n-1)+2
所以an是等差数列
S1=a1
所以2√a1=a1+1
(√a1-1)=0,an=1
an=2n-1
Sn=(1+2n-1)*n/2=n^2
bn=(-1)^n*n^2
Tn=b1+b2+...+bn
n为偶数,设n=2k.
Tn=-1^2+2^2-3^2+4^2+...-(2k-1)^2+(2k)^2
=(2-1)(2+1)+...+(2k-2k+1)(2k+2k-1)
=1+2+3+4+...+4k-1
=(1+4k-1)*2k/2
=4k^2
=n^2.
n是奇数时,设n=2k+1
Tn=4k^2+(-1)^(2k+1)*(2k+1)^2
=4k^2-(4k^2+4k+1)
=-4k-1
=-2(2k+1)+1
=-2n+1.

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