Question

已知f(x)=x+1,g(x)=2x+1，数列｛an}满足：a1=1,A(n+1)=f(an)(n为奇数）、g(an)(n为偶数），求an前2007项和

Answer 1

a(2n)=f[a(2n-1)]=a(2n-1)+1, a(2)=a(1)+1=2.
a(2n+1)=g[a(2n)]=2a(2n)+1=2[a(2n-1)+1] + 1 = 2a(2n-1) + 3,
a(2n+1)+3=2[a(2n-1)+3],
{a(2n-1)+3}是首项为a(1)+3=4,公比为2的等比数列.
a(2n-1)+3=4*2^(n-1)=2^(n+1).
a(2n-1)=2^(n+1)-3.

a(2n+2)=a(2n+1)+1=2a(2n)+1 + 1 = 2a(2n)+2,
a(2n+2)+2=2[a(2n)+2],
{a(2n)+2}是首项为a(2)+2=4,公比为2的等比数列.
a(2n)+2=4*2^(n-1)=2^(n+1).
a(2n)=2^(n+1)-2.

a(1)+a(2)+...+a(2007)=a(1)+a(3)+...+a(2*1004-1) + a(2)+a(4)+...+a(2*1003)
=2^2-3 + 2^3-3 + ... + 2^(1004+1)-3 + 2^2-2 + 2^3-2 + ... + 2^(1003+1)-2
=4[1+2+...+2^(1003)] -3*1004 + 4[1+2+...+2^(1002)]-2*1003
=4[2^(1004)-1]+4[2^(1003)-1]-3*1004-2*1003
=3*2^(1005) - 5028