Let \(\displaystyle 0<b<a\) and \(\displaystyle (x_{n})_{n\in \mathbb{N}}\) with\(\displaystyle x_{0}=1, \ x_{1}=a+b\)
\(\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}\)
a) If \(\displaystyle 0<b<a\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}\) then \(\displaystyle L= ?\)
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If \(\displaystyle 0<b<a<1\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}\) then \(\displaystyle L= ?\)
the right answer is \(\displaystyle L=\frac{1}{(1-a)(1-b)}\)
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far
\(\displaystyle x_{n+2}=(a+b)\cdot x_{n+1}-ab\cdot x_{n}\)
a) If \(\displaystyle 0<b<a\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\frac{x_{n+1}}{x_{n}}\) then \(\displaystyle L= ?\)
solutions to choose from 1. L=a 2. L=b 3. L=a/b 4. L=b/a 5. can't calculate ( right answer L=a)
b) If \(\displaystyle 0<b<a<1\) and \(\displaystyle L=\lim_{n\rightarrow \infty }\sum_{k=0}^{n}x_{k}\) then \(\displaystyle L= ?\)
the right answer is \(\displaystyle L=\frac{1}{(1-a)(1-b)}\)
I don't know how to start.I tried to write the first terms x1 x2... but I didn't get too far