1.已知数列{An}满足A(n+1)=2^n·An（n∈N*）,且A1=1则其通项公式An2.已知数列{Xn}满足递推公式{X(n+2)=X(n+1)-Xn、X1=a、X2=b},其中a,b为实常数,求X2009

Question

1.已知数列{An}满足A(n+1)=2^n·An （n∈N*）,且A1=1 则其通项公式An
2.已知数列{Xn}满足递推公式{X(n+2)=X(n+1)-Xn、X1=a、X2=b} ,其中a,b为实常数,求X2009

Answer 1

a(n+1)=2^n*a(n),
a(1)=1>0,a(2)>0,...,a(n)>0.
ln[a(n+1)] = nln(2) + ln[a(n)] = ln[a(n)] + ln(2)[n(n+1)-(n-1)n]/2,
ln[a(n+1)] - ln(2)*n(n+1)/2 = ln[a(n)] - ln(2)(n-1)n/2
{ln[a(n)]-ln(2)(n-1)n/2}是首项为ln[a(1)] - ln(2)*0=0的常数数列.
ln[a(n)]-ln(2)(n-1)n/2 = 0,
ln[a(n)] = ln(2)(n-1)n/2 = ln{2^[(n-1)n/2]},
a(n) = 2^[(n-1)n/2].
x(n+2)=x(n+1)-x(n),
x(1)=a,
x(2)=b,
x(3)=b-a,
x(4)=-a,
x(5)=-b,
x(6)=a-b,
x(7)=a,
x(8)=b,
x(9)=b-a,
...
x(n+6)=x(n)
{x(n)}是周期为6的数列.
x(2009)=x(6*333+1)=x(1)=a