已知{an}是公比为正数的等比数列,且1/a2+1/a3+1/a4=117,a1*a2*a3=1/3^6,求 lim(a1+a2+a3+.+an)
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![已知{an}是公比为正数的等比数列,且1/a2+1/a3+1/a4=117,a1*a2*a3=1/3^6,求 lim(a1+a2+a3+.+an)](/uploads/image/z/7101023-23-3.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E6%98%AF%E5%85%AC%E6%AF%94%E4%B8%BA%E6%AD%A3%E6%95%B0%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%941%2Fa2%2B1%2Fa3%2B1%2Fa4%3D117%2Ca1%2Aa2%2Aa3%3D1%2F3%5E6%2C%E6%B1%82+lim%28a1%2Ba2%2Ba3%2B.%2Ban%29)
已知{an}是公比为正数的等比数列,且1/a2+1/a3+1/a4=117,a1*a2*a3=1/3^6,求 lim(a1+a2+a3+.+an)
已知{an}是公比为正数的等比数列,且1/a2+1/a3+1/a4=117,a1*a2*a3=1/3^6,求 lim(a1+a2+a3+.+an)
已知{an}是公比为正数的等比数列,且1/a2+1/a3+1/a4=117,a1*a2*a3=1/3^6,求 lim(a1+a2+a3+.+an)
因为a1*a2*a3=1/3^6,所以a2^3=1/3^6,所以a2=1/9
1/a2+1/a3+1/a4=(1+1/q+1/q^2)/a2=117,所以(1+1/q+1/q^2)=13解得q=1/3(负值舍去),所以a1=1/3
所以a1+a2+a3+.+an=a1(1-q^n)/(1-q)=1/3*(1-1/3^n)/(1-1/3)=(1-1/3^n)/2
lim(a1+a2+a3+.+an)=1/2
1/a2+1/a3+1/a4=117 则 1/a2 *(1+1/q +1/q^2) =117 (1)
a1*a2*a3=1/3^6 则 a2^3 = 1/3^6 a2 = 1/9 或 -1/9
1/a2 *(1+1/q +1/q^2) =117 >0 且q>0 则a2 ...
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1/a2+1/a3+1/a4=117 则 1/a2 *(1+1/q +1/q^2) =117 (1)
a1*a2*a3=1/3^6 则 a2^3 = 1/3^6 a2 = 1/9 或 -1/9
1/a2 *(1+1/q +1/q^2) =117 >0 且q>0 则a2 取 1/9
带入 (1) (1+1/q +1/q^2) = 117/9 =13 则q=1/3 或 -1/4 (舍)
a2 = 1/9 q= 1/3
lim(a1+a2+a3+。。。+an)
= lim[1/3 * (1- (1/3)^n)/ (1- 1/3)]
= 1/2
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