作业帮设数列的前n项和为sn,且Sn=n^2-6n+c(c∈R).
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设数列的前n项和为sn,且Sn=n^2-6n+c(c∈R).
设数列的前n项和为sn,且Sn=n^2-6n+c(c∈R).
设数列的前n项和为sn,且Sn=n^2-6n+c(c∈R).
1.
c=1 Sn=n²-6n+1
n=1时,a1=S1=1²-6×1+1=-4
n≥2时,an=Sn-S(n-1)=n²-6n+1-[(n-1)²-6(n-1)+1]=2n-7
n=1时,a1=2×1-7=-5≠-4
数列{an}的通项公式为
an=-4 n=1
2n-7 n≥2
2.
c=0 Sn=n²-6n
n=1时,a1=S1=1²-6×1=-5
n≥2时,an=Sn-S(n-1)=n²-6n-[(n-1)²-6(n-1)]=2n-7
n=1时,a1=2×1-7=-5,同样满足通项公式
数列{an}的通项公式为an=2n-7
bn=an/2ⁿ=(2n-7)/2ⁿ
n>1
Tn=b1+b2+...+bn=(2×1-7)/2 +(2×2-7)/2²+(2×3-7)/2³+...+(2n-7)/2ⁿ
Tn /2=(2×1-7)/2²+(2×2-7)/2³+...+[2(n-1)-7]/2ⁿ+(2n-7)/2^(n+1)
Tn-Tn /2=Tn /2=(-5/2) +2/2²+2/2³+...+2/2ⁿ -(2n-7)/2^(n+1)
=(-5/2) +2(1/2²+1/2³+...+1/2ⁿ) -(2n-7)/2^(n+1)
=(-5/2) +2×(1/4)×[1-1/2^(n-1)]/(1-1/2) -(2n-7)/2^(n+1)
=(-3/2) -2/2ⁿ -(2n-7)/2^(n+1)
Tn=-3- (2n-3)/2ⁿ
n>1 n=2时,(2n-3)/2ⁿ=(2×2-3)/2²=1/4 T2=-3-1/4=-13/4
n=3时,(2n-3)/2ⁿ=(2×3-3)/2³=3/8>1/4 T3=-3- 3/8=-27/8
n≥3时,
{[2(n+1)-3]/2^(n+1)}/[(2n-3)/2ⁿ]=(2n-1)/(4n-6)
n≥3 (2n-1)-(4n-6)=5-2n0 Tn->-3
综上,得Tn的取值范围为[-27/8,-3)