高二数列问题。急急急·~~~求高手帮忙!
等比数列an的前n项和为Sn,首项a1=a,公比为q,计算liman/Sn的值2求数列0.18,0.0018,0.000018…(18上面是循环)的前n项和及各项和。...
等比数列an的前n项和为Sn,首项a1=a,公比为q,计算liman/Sn的值
2 求数列0.18,0.0018,0.000018…(18上面是循环)的前n项和及各项和。 展开
2 求数列0.18,0.0018,0.000018…(18上面是循环)的前n项和及各项和。 展开
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Sn=a1+a1q+a1q²…
qSn=a1q+a1q²…+a1q^n
(1-q)=a1-a1q^n
得到:
Sn=(a1-a1q^n)/(1-q)=a(1-q^n)/(1-q)
liman/Sn=aq^(n-1)/[a(1-q^n)/(1-q)]=q^(n-1)(1-q)/(1-q^n)
=(1-q)q^(n-1)/(1-q^n)
=[q^(n-1)-q^n]/(1-q^n)
-1<q<1 为 0
q>1或者q<-1时 1-1/q
q=1 为0
q=-1 没有
0.18,0.0018,0.000018…(18上面是循环)的前n项和及各项和。
A1=0.18 18上面是循环
A2=0.0018 18上面是循环
A3=0.000018 18上面是循环
An是等比数列,公比为1/100
An=0.18 *(1/100 ) ^(n-1 ) 18上面是循环
Sn=A1*[1-(1/00)^n]/(1-q)
A1自身也是等比数列的前n项和
A1=0.18/(1-1/100)=2/11
An=2/11 *(1/100 ) ^(n-1 )
Sn=2/11*[1-(1/00)^n]/(1-q)
qSn=a1q+a1q²…+a1q^n
(1-q)=a1-a1q^n
得到:
Sn=(a1-a1q^n)/(1-q)=a(1-q^n)/(1-q)
liman/Sn=aq^(n-1)/[a(1-q^n)/(1-q)]=q^(n-1)(1-q)/(1-q^n)
=(1-q)q^(n-1)/(1-q^n)
=[q^(n-1)-q^n]/(1-q^n)
-1<q<1 为 0
q>1或者q<-1时 1-1/q
q=1 为0
q=-1 没有
0.18,0.0018,0.000018…(18上面是循环)的前n项和及各项和。
A1=0.18 18上面是循环
A2=0.0018 18上面是循环
A3=0.000018 18上面是循环
An是等比数列,公比为1/100
An=0.18 *(1/100 ) ^(n-1 ) 18上面是循环
Sn=A1*[1-(1/00)^n]/(1-q)
A1自身也是等比数列的前n项和
A1=0.18/(1-1/100)=2/11
An=2/11 *(1/100 ) ^(n-1 )
Sn=2/11*[1-(1/00)^n]/(1-q)
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