Sn为数列的前n项和,Sn=-1的n次方乘以an-2的n次方分之1,求S1+...+S100
1个回答
n=1时,a1=S1=(-1)×a1- 1/2
a1=-1/4
n≥2时,an=Sn-S(n-1)=(-1)ⁿ×an -1/2ⁿ -(-1)^(n-1)×a(n-1)+1/2^(n-1)
an=(-1)ⁿ×an +(-1)ⁿ×a(n-1)+1/2ⁿ
n≥2时,
n为奇数时,2an=-a(n-1)+1/2ⁿ
2an-2/2^(n+2)=-a(n-1)+1/2^(n+1)
[an-1/2^(n+2)]/[a(n-1)-1/2^(n+1)]=-1/2,为定值.
a1-1/2³=-1/4-1/8=-3/8,数列{an -1/2^(n+2)}是以-3/8为首项,-1/2为公比的等比数列.
a(n-1)-1/2^(n+1)=(-3/8)×(-1/2)^(n-2)=3/2^(n+1)
a(n-1)=1/2^(n-1)
n为偶数时,a(n-1)=-1/2ⁿ
即：n为奇数时,an=-1/2^(n+1)；n为偶数时,an=1/2ⁿ
n为奇数时,Sn=(-1)ⁿ×an- 1/2ⁿ=-1/2^(n+1)
n为偶数时,Sn=(-1)ⁿ×an- 1/2ⁿ=0
S1+S2+...+S100
=(S1+S3+...+S99)+(S2+S4+...+S100)
=[(-1/2²)+(-1/2⁴)+...+(-1/2)^100]+0
=(-1/4)×[1-(1/4)^50]/(1-1/4)
=(-3)×(1- 1/4^50)
=-3 +3/4^50
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