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

题目

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

答案

bn=a(2n+1)+4n-2b(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=-...