設(shè) A(n+1)-a*A(n)=b*(A((n)-a*A(n-1))
=>a+b=14a*b=1=>a=7-4*3^0.5b=7+4*3^0.5
=> A(n)-a*A(n-1)=b^(n-1)*(A(1)-a*A(0))A1=A0=1=>A(n)-a*A(n-1)=b^(n-1)*(1-a)
設(shè)A(n)+c*b^n=a*(A(n-1)+c*b^(n-1))=>A(n)=a*A(n-1)+a*c*b^(n-1)-c*b^(n)=>c=(1-a)/(a-b)
=>A(n)+c*b^n=a^n*(A0+c) =>An=a^n*(1+c)-c*b^n=(-(1-a)*b^n+(1-b)*a^n)/(a-b)
=>2*A(n)-1=2*(-(1-a)*b^n+(1-b)*a^n)/(a-b)-1ab=1 a+b=14
=>2*A(n)-1=2*((a-1)*b^n+(1-b)*b^n)/(a-b)-1
=2*{a*b^n-b^n+a^n-a^n*b}/(a-b)-1
=(2*a^n - a + 2*a*(1/a)^n - 1)/(a + 1)
化簡(jiǎn) ,或用數(shù)學(xué)歸納法就可證明