Question

已知数列log2(an-1)为等差数列且a1=3 a2=5

Answer 1

设数列 log2(an-1) 公差为d

d = long2(an-1)-log2(a(n-1)-1) = log2[ (an-1)/(a(n-1)-1]

所以 (an-1)/(a(n-1)-1) = 2^d

而由a1=3 a2=5,可知 d=log2 4/long2 2 = 2

所以,an-1 = 2^n

2.a(k+1) -ak = 2^(k+1)-2^k =2^k

所以 a2-a1）分之一+（a3-a2）分之一+.+（（an+1）—an）分子一

= 1/2+1/2^2+...+1/2^n =1-1/2^n