设f(λ)=λ/(1+λ)(λ≠-1,0).数列{bn}满足b1=1/2,bn=f(bn-1)注bn-1 n-1是角标n
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设f(λ)=λ/(1+λ)(λ≠-1,0).数列{bn}满足b1=1/2,bn=f(bn-1)注bn-1 n-1是角标n≥2
求证数列{1/bn}为等差 此问已求1/bn=n+1
记cn=(1/2)^(n-1)乘以(1/bn-1),数列{cn}的前项和为Tn,求证当n≥2时,3≤Tn+cn
求证数列{1/bn}为等差 此问已求1/bn=n+1
记cn=(1/2)^(n-1)乘以(1/bn-1),数列{cn}的前项和为Tn,求证当n≥2时,3≤Tn+cn
![设f(λ)=λ/(1+λ)(λ≠-1,0).数列{bn}满足b1=1/2,bn=f(bn-1)注bn-1 n-1是角标n](/uploads/image/z/5057204-68-4.jpg?t=%E8%AE%BEf%28%CE%BB%29%3D%CE%BB%2F%281%2B%CE%BB%29%28%CE%BB%E2%89%A0-1%2C0%EF%BC%89.%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E6%BB%A1%E8%B6%B3b1%3D1%2F2%2Cbn%3Df%28bn-1%29%E6%B3%A8bn-1+n-1%E6%98%AF%E8%A7%92%E6%A0%87n)
Cn = n* (1/2)^(n-1),由错位相减法求Tn:
Tn = 1 + 2* (1/2) + 3* (1/2)²+ 4* (1/2)³ +...+ n* (1/2)^(n-1),——①
1/2 * Tn = 1/2 + 2* (1/2)²+ 3 (1/2)³ +...+ (n-1)* (1/2)^(n-1) + n* (1/2)^n,——②
①-②,得:1/2 * Tn = 1+ 1/2 + (1/2)² + (1/2)³+...+(1/2)^(n-1) - n* (1/2)^n,
求得 Tn = 4 - (2n+4)* (1/2)^n
所以 Tn + Cn = 4 - (2n+4)* (1/2)^n + (2n)* (1/2)^n = 4 - 4* (1/2)^n = 4*{1- (1/2)^n},
显然 Tn + Cn < 4,
而由于当n≥2时,(1/2)^n ≤ 1/4 ,得证 Tn + Cn ≥ 3 .
Tn = 1 + 2* (1/2) + 3* (1/2)²+ 4* (1/2)³ +...+ n* (1/2)^(n-1),——①
1/2 * Tn = 1/2 + 2* (1/2)²+ 3 (1/2)³ +...+ (n-1)* (1/2)^(n-1) + n* (1/2)^n,——②
①-②,得:1/2 * Tn = 1+ 1/2 + (1/2)² + (1/2)³+...+(1/2)^(n-1) - n* (1/2)^n,
求得 Tn = 4 - (2n+4)* (1/2)^n
所以 Tn + Cn = 4 - (2n+4)* (1/2)^n + (2n)* (1/2)^n = 4 - 4* (1/2)^n = 4*{1- (1/2)^n},
显然 Tn + Cn < 4,
而由于当n≥2时,(1/2)^n ≤ 1/4 ,得证 Tn + Cn ≥ 3 .
设f(λ)=λ/(1+λ)(λ≠-1,0).数列{bn}满足b1=1/2,bn=f(bn-1)注bn-1 n-1是角标n
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