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

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
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

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
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)+a(3)+...+2a(n-1)+a(n-1)+a(n)+a(n)+a(n+1)
=na(1)+(n-1)a(2)+...+2a(n-1)+a(n)+[a(1)+a(2)+...+a(n)+a(n+1)]
=b(n)+[1+2+...+2^(n-1)+2^n]
=b(n)+[2^(n+1)-1]/[2-1]
=b(n)+2^(n+1)-1,
b(n+1)-2^(n+2)=b(n)-2^(n+1)-1=b(n)-2^(n+1)-(n+1-n),
b(n+1)-2^(n+1+1)+(n+1)=b(n)-2^(n+1)+n,
{b(n)-2^(n+1)+n}是各项为b(1)-2^(1+1)+1=-2的常数数列.
b(n)-2^(n+1)+n=b(1)-2^(1+1)+1=-2,
b(n)=2^(n+1)-n-2,
b(2n-1)=2^(2n-1+1)-(2n-1)-2=4^n-2n-1
b(1)+b(3)+...+b(2n-1)=[4+4^2+...+4^n]-2[1+2+...+n]-n
=4[1+4+...+4^(n-1)]-n(n+1)-n
=4[4^n-1]/(4-1)-n^2-2n
=[4^(n+1)-4]/3 - n^2 - 2n