Question

若数列{an}的通项公式是an=3^-n+2^-n+(-1)^n(3^-n-2^-n)]/2,n=1,2,…,

则limn→∞（a1+a2+…+an)等于（ ）

A.11/24 B.17/24 C.19/24 D.25/24

Answer 1

an=[ 3^(-n)+2^(-n)+(-1)^n. (3^(-n)-2^(-n)) ]/2
when n is odd
an = 2^(-n)
a1+a3+...+a(2n-1) = 2^(-1) +2^(-3)+...+ 2^(-(2n-1))
= (1/3)(1- 2^(-2n))
when n is even

an = 3^(-n)
a2+a4+...+a2n = 3^(-2)+3^(-4)+...+3^(-2n)
= (1/8) (1- 3^(-2n) )

a1+a2+...+a(2n) = (1/3)[1- 2^(-2n)] + (1/8) (1- 3^(-2n) )
lim(2n->∞) (a1+a2+..+a2n) = 1/3+ 1/8 = 11/24

lim(n->∞) (a1+a2+..+an) = 11/24
Ans: A:11/24