高中数列难题已知数列an,bn满足:a1=9/2,2a(下标n+1)-an=6*2^n,bn=an-2^(n+1),那么记数列an,bn的前n项和分别为Sn,Tn,若对任意的nεn+都有Sn/Tn≦m/bn,求实数m的最小值.高考在即,加油吧
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![高中数列难题已知数列an,bn满足:a1=9/2,2a(下标n+1)-an=6*2^n,bn=an-2^(n+1),那么记数列an,bn的前n项和分别为Sn,Tn,若对任意的nεn+都有Sn/Tn≦m/bn,求实数m的最小值.高考在即,加油吧](/uploads/image/z/15209113-49-3.jpg?t=%E9%AB%98%E4%B8%AD%E6%95%B0%E5%88%97%E9%9A%BE%E9%A2%98%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%2Cbn%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D9%2F2%2C2a%28%E4%B8%8B%E6%A0%87n%2B1%29-an%3D6%2A2%5En%2Cbn%3Dan-2%5E%28n%2B1%29%2C%E9%82%A3%E4%B9%88%E8%AE%B0%E6%95%B0%E5%88%97an%2Cbn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E4%B8%BASn%2CTn%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%CE%B5n%EF%BC%8B%E9%83%BD%E6%9C%89Sn%EF%BC%8FTn%E2%89%A6m%EF%BC%8Fbn%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.%E9%AB%98%E8%80%83%E5%9C%A8%E5%8D%B3%2C%E5%8A%A0%E6%B2%B9%E5%90%A7)
高中数列难题已知数列an,bn满足:a1=9/2,2a(下标n+1)-an=6*2^n,bn=an-2^(n+1),那么记数列an,bn的前n项和分别为Sn,Tn,若对任意的nεn+都有Sn/Tn≦m/bn,求实数m的最小值.高考在即,加油吧
高中数列难题
已知数列an,bn满足:a1=9/2,2a(下标n+1)-an=6*2^n,bn=an-2^(n+1),那么记数列an,bn的前n项和分别为Sn,Tn,若对任意的nεn+都有Sn/Tn≦m/bn,求实数m的最小值.
高考在即,加油吧
高中数列难题已知数列an,bn满足:a1=9/2,2a(下标n+1)-an=6*2^n,bn=an-2^(n+1),那么记数列an,bn的前n项和分别为Sn,Tn,若对任意的nεn+都有Sn/Tn≦m/bn,求实数m的最小值.高考在即,加油吧
2a(n+1) -an=6×2^n
2a(n+1)=an+6×2^n
2a(n+1)-2×2^(n+2)=an-2^(n+1)
[a(n+1)-2^(n+2)]/[an-2^(n+1)]=1/2,为定值.
a1-2^2=9/2 -4=1/2
数列{an -2^(n+1)}是以1/2为首项,1/2为公比的等比数列.
an -2^(n+1)=1/2ⁿ
an=2^(n+1) +1/2ⁿ
n=1时,a1=2^2 +1/2=9/2,同样满足.
数列{an}的通项公式为an=2^(n+1) +1/2ⁿ.
bn=an-2^(n+1)=2^(n+1)+1/2ⁿ-2^(n+1)=1/2ⁿ
Sn=a1+a2+...+an=2^2+2^3+...+2^(n+1)+1/2^1+1/2^2+...+1/2ⁿ
=4(2ⁿ -1)/(2-1) +(1/2)(1-1/2ⁿ)/(1-1/2)
=2^(n+2) -1/2ⁿ -3
Tn=b1+b2+...+bn=1/2+1/2^2+...+1/2ⁿ=(1/2)(1-1/2ⁿ)/(1-1/2)=1- 1/2ⁿ
Sn/Tn≤m/bn
[2^(n+2) -1/2ⁿ -3]/(1-1/2ⁿ)≤m/(1/2ⁿ)
2^(n+2) -1/2ⁿ -3≤m(2ⁿ -1)
m≥[2^(n+2) -1/2ⁿ -3]/(2ⁿ -1)
m≥[2^(2n+2) -3×2ⁿ -1]/[2^(2n) -2ⁿ]
m≥[4×2^(2n)-4×2ⁿ+2ⁿ -1]/[2^(2n) -2ⁿ]
m≥4+(2ⁿ -1)/[2ⁿ(2ⁿ -1)]
m≥4 +1/2ⁿ
随n增大,2ⁿ递增,1/2ⁿ递减,4+1/2ⁿ递减,因此当n=1时,4+1/2ⁿ有最大值4+1/2=9/2
要对任意正整数n,不等式恒成立,则m≥9/2
m的最小值为9/2.