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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/25 19:46:11
![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](/uploads/image/z/693518-14-8.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%E2%89%A52%E6%B1%82%E8%AF%81%E6%95%B0%E5%88%97%EF%BD%9B1%2Fbn%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE+%E6%AD%A4%E9%97%AE%E5%B7%B2%E6%B1%821%2Fbn%3Dn%2B1%E8%AE%B0cn%3D%281%2F2%29%5E%28n-1%29%E4%B9%98%E4%BB%A5%281%2Fbn-1%29%2C%E6%95%B0%E5%88%97%EF%BD%9Bcn%EF%BD%9D%E7%9A%84%E5%89%8D%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82%E8%AF%81%E5%BD%93n%E2%89%A52%E6%97%B6%2C3%E2%89%A4Tn%2Bcn)
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
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
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
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 .