Question

高中数列数学题求解

如图

Answer 1

A2 - A3 = A2 * (1-q)
A3 - A4 = A3 * (1-q)
既然 A2 - A3 = 2(A3 - A4)
所以 A2 * (1 - q) = 2A3 * (1 - q)
因此，A2 = 2 A3
所以，q = A3/A2 = 1/2
那么，An = A1 * (1/2)^(n-1) = 64 * (1/2)^(n-1) = 128 * (1/2)^n = 2^(7-n)
那么就有：
Bn = 7 - n
因此
Tn = 7n - (1+2+……+n)
= 7n - n(n+1)/2
= (14n - n² - n)/2
= n(14 - n)/2

Answer 2

(I)
an=a1.q^(n-1)
a2-a3=2(a3-a4)
a1.q(1-q) = 2a1q^2.(1-q)
2q^2 -q =0
q(2q-1)=0
q=1/2
ie
an = 64.(1/2)^(n-1) = (1/2)^(n-7)
(II)
bn=log<2>an = -(n-7)
n≤7
Tn=n(13-n)/2
n>7
Tn

=T7 +|b8|+|b9|+...+|bn|
=7(13-7)/2 +|b8|+|b9|+...+|bn|
=21 -(b8+b9+....+bn)
=21+ (1/2)(n-7 +1)(n-8)
=21+ (1/2)(n-6)(n-8)