设数列an为等比数列,数列bn满足bn=na1+(n-1)a2+...+2an-1+an已知b1=1,b2=4
问题描述：
设数列an为等比数列,数列bn满足bn=na1+(n-1)a2+...+2an-1+an已知b1=1,b2=4
求(1)数列｛an｝的首项和公比.(2)bn.(3)b1+b3+b5+b7+...+b2n-1
答
1=b(1)=1*a(1)=a(1),4=b(2)=2a(1)+a(2), 2=a(2).q=a(2)/a(1)=2,a(n)=a(1)*q^(n-1)=1*2^(n-1)=2^(n-1)b(n+1)=(n+1)a(1)+(n+1-1)a(2)+(n+1-2)a(3)+...+3a(n+1-2)+2a(n+1-1)+a(n+1)=na(1)+a(1)+(n-1)a(2)+a(2)+(n-2)a(3...b(n+1)-2^(n+1+1)+(n+1)=b(n)-2^(n+1)+n,最后的+n哪里来的？？b(n+1)=b(n)+2^(n+1)-1 = b(n) + 2^(n+1) - (n+1-n),b(n+1)+(n+1)=b(n) + 2^(n+1) +n