已知数列an满足条件,a1=2,a2=3,2an+1=3an-an-1(n大于等于2证明an-1已知数列an满足条件,a1=2,a2=3,2an+1=3an-an-1(n大于等于2证明an-1-an为等比数列求不等式an+1-m分之an-m小于三分之二,成立的所有正整数m,n的

Question

已知数列an满足条件,a1=2,a2=3,2an+1=3an-an-1(n大于等于2 证明an-1
已知数列an满足条件,a1=2,a2=3,2an+1=3an-an-1(n大于等于2
证明an-1-an为等比数列
求不等式an+1-m分之an-m小于三分之二,成立的所有正整数m,n的值

Answer 1

2a(n+1)=3a(n) - a(n-1),
2a(n+2) = 3a(n+1) - a(n),
2a(n+2)-2a(n+1) = a(n+1) - a(n),
a(n+2) - a(n+1) = [a(n+1)-a(n)]/2,
{a(n+1)-a(n)}是首项为a(2)-a(1)=1,公比为(1/2)的等比数列.
a(n+1)-a(n) = (1/2)^(n-1) = 1/2^(n-1).
2a(n+2)-a(n+1) = 2a(n+1)-a(n),
{2a(n+1)-a(n)}是首项为2a(2)-a(1)=4,的常数数列.
2a(n+1)-a(n) = 4,
a(n+1) - a(n)/2 = 2.
1/2^(n-1) - 2 = [a(n+1)-a(n)] - [a(n+1)-a(n)/2] = -a(n)/2.
a(n) = 4 - 1/2^(n-2).
n-m>=1时,
a(n-m) = 4 - 1/2^(n-m-2).
a(n-m+1) = 4 - 1/2^(n-m-1).
2/3 > a(n-m)/a(n-m+1) = [4-1/2^(n-m-2)]/[4-1/2^(n-m-1)] = [2^(n-m+1) - 2]/[2^(n-m+1)-1],
2[2^(n-m+1)-1] > 3[2^(n-m+1)-2],
2^(n-m+1) < 4 = 2^2.
2= 1+1