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-

Answer 1

n=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