已知数列an满足a1=2/5,且对任意n属于N,都有an/a(n+1)=4an+2/a(n+1)+2令bn=anan

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/26 10:58:34

已知数列an满足a1=2/5,且对任意n属于N*,都有an/a(n+1)=4an+2/a(n+1)+2令bn=an*an+1,Tn=b1+b2+b3.+bn求证Tn小于4|15

a(n)/a(n+1)=[4a(n)+2]/[a(n+1)+2],
[a(n+1)+2]/a(n+1)=[4a(n)+2]/a(n),
2/a(n+1) + 1 = 2/a(n) + 4,
2/a(n+1) = 2/a(n) + 3,
{2/a(n)}是首项为2/a(1)=5, 公差为3的等差数列.
2/a(n) = 5 + 3(n-1) = 3n + 2,
a(n) = 2/(3n+2),
b(n) = a(n)a(n+1) = 4/[(3n+2)(3n+5)] = (4/3)[1/(3n+2) - 1/(3n+5)] = (4/3) [1/(3n+2) - 1/(3(n+1)+2)],
t(n) = b(1)+b(2) + ... + b(n-1)+b(n)
=(4/3)[1/(3*1+2) - 1/(3*2+2) + 1/(3*2+2)-1/(3*3+2) + ... + 1/[3(n-1)+2]-1/(3n+2) + 1/(3n+2) - 1/[3(n+1)+2] ]
=(4/3)[1/(3*1+2) - 1/[3(n+1)+2] ]
< (4/3)[ 1/5]
= 4/15

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