Question

已知数列{an}满足:a1=1,an+1=1/2an+n,n 为奇数,an-2n,n 为偶数.设bn=a2n+1+4n-2,n€N

求证：数列{bn}是等比数列，并求其通项公式。(2)求数列an的前100项中，所有奇数项的和S

Answer 1

bn=a(2n+1)+4n-2
b(n+1)=a(2n+3)+4n+2
=a(2n+2)-2(2n+2)+4n+2
=1/2a(2n+1)+2n-1
=1/2[a(2n+1)+4n-2]
∴b(n+1)/bn=1/2
∴数列{bn}是等比数列,公比为1/2
b1=a3+2=a2-4+2=1/2a1+1-2=-1/2
bn=-(1/2)^n
2
∵bn=a(2n+1)+4n-2
∴a(2n+1)=bn-4n+2=-1/2^n-4n+2
S=a1+a3+a5+....+a99
=1+(-1/2-1/4-1/8-...-1/2^49)-4(1+2+3+...+49)+2·49
=1-(1-1/2^49)-2*49*50+98
= 1/2^49-4802