已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q(q不等于1) 的等比
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已知函数f(x)=(x-1)^2,数列an是公差为d的等差数列,bn是公比为q(q不等于1) 的等比
(2008•丰台区一模)已知函数f(x)=(x-1)2,数列{an}是公差为d的等差数列,{bn}是公比为q(q∈R,q≠1)的等比数列.若a1=f(d-1),a3=f(d+1),b1=f(q-1),b3=f(q+1).
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅲ)试比较
3bn−1 /3bn+1 与an+1/ an+2
的大小. an=2(n-1).bn=3n-1.
就是想问一下最后一问如果用二项式定理怎么证明
(2008•丰台区一模)已知函数f(x)=(x-1)2,数列{an}是公差为d的等差数列,{bn}是公比为q(q∈R,q≠1)的等比数列.若a1=f(d-1),a3=f(d+1),b1=f(q-1),b3=f(q+1).
(Ⅰ)求数列{an},{bn}的通项公式;
(Ⅲ)试比较
3bn−1 /3bn+1 与an+1/ an+2
的大小. an=2(n-1).bn=3n-1.
就是想问一下最后一问如果用二项式定理怎么证明
(I)
f(x)=(x-1)^2
an=a1+(n-1)d
bn =b1q^(n-1) ; q≠1
a1=f(d-1)
=(d-2)^2 (1)
a3= f(d+1)
a1+2d = d^2 (2)
(2)-(1)
2d= 4d-4
d= 2
a1= 0
an = 2n-2
b1=f(q-1)
= (q-2)^2 (3)
b3=f(q+1)
b1q^2 = q^2
b1=1
=> q=3
bn = 3^(n-1)
(III)
(3bn−1)/(3bn+1) 与(an+1)/( an+2)
(3bn−1)/(3bn+1) = [3^(n-1) - 1]/[3^(n-1) + 1]
= 1 - 2/[3^(n-1) + 1]
(an+1)/( an+2) = (2n-2+1)/(2n-2+2)
= (2n-1)/(2n)
= 1- 1/(2n)
(an+1)/( an+2) > (3bn−1)/(3bn+1)
f(x)=(x-1)^2
an=a1+(n-1)d
bn =b1q^(n-1) ; q≠1
a1=f(d-1)
=(d-2)^2 (1)
a3= f(d+1)
a1+2d = d^2 (2)
(2)-(1)
2d= 4d-4
d= 2
a1= 0
an = 2n-2
b1=f(q-1)
= (q-2)^2 (3)
b3=f(q+1)
b1q^2 = q^2
b1=1
=> q=3
bn = 3^(n-1)
(III)
(3bn−1)/(3bn+1) 与(an+1)/( an+2)
(3bn−1)/(3bn+1) = [3^(n-1) - 1]/[3^(n-1) + 1]
= 1 - 2/[3^(n-1) + 1]
(an+1)/( an+2) = (2n-2+1)/(2n-2+2)
= (2n-1)/(2n)
= 1- 1/(2n)
(an+1)/( an+2) > (3bn−1)/(3bn+1)
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