数列满足a1=3,a2=7,an=a(n-1)+a(n-2)+a(n-1)a(n-2),则a10=快

Question

数列满足a1=3,a2=7,an=a(n-1)+a(n-2)+a(n-1)a(n-2),则a10=
快

Answer 1

a(n) + 1 = a(n-1) + a(n-2) + a(n-1)a(n-2) + 1 = [a(n-1)+1][a(n-2)+1],
b(n) = a(n) + 1,b(1)=a(1)+1=4,b(2)=a(2)+1=7+1=8.
b(n+2)=b(n+1)*b(n),
c(n) = ln[b(n)],
c(n+2) = c(n+1) + c(n),
c(n+2)+xc(n+1) = [1+x][c(n+1)+xc(n)],1 = x(1+x),0 = x^2 + x - 1,x=(5^(1/2)-1)/2.
{c(n+1)+xc(n)}是首项为c(2)+xc(1)=ln[b(2)]+xln[b(1)]=ln[a(2)+1]+xln[a(1)+1]=ln4+xln8=[2 + 3x]ln2,公比为(1+x)的等比数列.
c(n+1)+xc(n)=[2+3x]ln2*(1+x)^(n-1).
x^(n)c(n+1) + x^2*x^(n-1)c(n) = x(2+3x)ln2*[x(1+x)]^(n-1) = x(2+3x)ln2,
d(n) = x^(n-1)c(n),
d(n+1) + x^2d(n) = x(2+3x)ln2,
d(n+1) + y = -x^2[d(n) + y],x(2+3x)ln2=-y(x^2+1),y=-x(2+3x)ln2/(x^2+1).
{d(n)+y}是首项为d(1)+y=c(1)+y=ln[b(1)]+y=ln4+y,公比为-x^2的等比数列.
d(n)+y=[ln4+y](-x^2)^(n-1),
c(n) = d(n)(1+x)^(n-1) = {(ln4+y)(-x^2)^(n-1)-y}(1+x)^(n-1)
=(ln4+y)(-x)^(n-1) - y(1+x)^(n-1),
b(n) = e^[c(n)] = e^{(ln4+y)(-x)^(n-1) - y(1+x)^(n-1)},
a(n) = b(n)-1 = e^{(ln4+y)(-x)^(n-1) - y(1+x)^(n-1)} - 1
a(10) = e^{(ln4+y)(-x)^9 - y(1+x)^9}-1,
y = -x(2+3x)ln2/(x^2+1),
x = [5^(1/2)-1]/2