等差数列 等比数列 {an}是等差数列,{bn}是各项为正的等比数列,且a1=b1=1,a3+b5=21,a5+b3=1
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等差数列 等比数列
{an}是等差数列,{bn}是各项为正的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13
(1)求{an},{bn} 通向公式
(2)求数列{an/bn}的前n项和
{an}是等差数列,{bn}是各项为正的等比数列,且a1=b1=1,a3+b5=21,a5+b3=13
(1)求{an},{bn} 通向公式
(2)求数列{an/bn}的前n项和
![等差数列 等比数列 {an}是等差数列,{bn}是各项为正的等比数列,且a1=b1=1,a3+b5=21,a5+b3=1](/uploads/image/z/1304745-33-5.jpg?t=%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97+%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%7Bbn%7D%E6%98%AF%E5%90%84%E9%A1%B9%E4%B8%BA%E6%AD%A3%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E4%B8%94a1%3Db1%3D1%2Ca3%2Bb5%3D21%2Ca5%2Bb3%3D1)
(1)a1+2d+b1×q^4=21 ①
a1+4d+b1×q^2=13 ②
①×2-②=2q^4-q^2=28
解得q=2
∴d=2
∴an=2n-1
bn=2^(n-1)
(2)an/bn=2n×2^(1-n)-2^(1-n)
∴Sn=2×1-1+4×1/2-1/2+8×1/4-1/4+…+2n×2^(1-n)-2^(1-n)
=2n-(1/2+1/4+…+2^(1-n))
=2n-(1/2)^n+1
这个q^4是表示q的四次方,下同
给分吧,打这种数学符号真他妈辛苦
a1+4d+b1×q^2=13 ②
①×2-②=2q^4-q^2=28
解得q=2
∴d=2
∴an=2n-1
bn=2^(n-1)
(2)an/bn=2n×2^(1-n)-2^(1-n)
∴Sn=2×1-1+4×1/2-1/2+8×1/4-1/4+…+2n×2^(1-n)-2^(1-n)
=2n-(1/2+1/4+…+2^(1-n))
=2n-(1/2)^n+1
这个q^4是表示q的四次方,下同
给分吧,打这种数学符号真他妈辛苦
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