Answer:
(a) [tex]L_{n}=\frac{1}{2}L_{n-1}+\frac{1}{2}L_{n-2}[/tex]
(b) [tex]a_{n}=266666.67(-\frac{1}{2})^{n}+233333.33[/tex]
Explanation:
(a) Recurrence relation for {Ln}
[tex]L_{n}=\frac{1}{2}(L_{n-1}+L_{n-2})\\\\L_{n}=\frac{1}{2}L_{n-1}+\frac{1}{2}L_{n-2}[/tex]
(b)
[tex]r^{2}-\frac{1}{2}r-\frac{1}{2}=0\\ \\\frac{1}{2}(2r+1)(r-1)=0\\\\(1) 2r+1=0 --> r=-\frac{1}{2}\\\\(2)r-1=0--> r=1[/tex]
General solution is:
[tex]a_{n}=k_{1}(-\frac{1}{2} )^{n}+k_{2}[/tex]
Considering initial conditions:
[tex](a)...(-\frac{1}{2})k_{1}+k_{2}=100000\\\\(b)...(\frac{1}{4})k_{1}+k_{2}=300000[/tex]
Solving the equations:
[tex]k_{1}=\frac{800000}{3}=266666.67\\ \\k_{2}=\frac{700000}{3}=233333.33\\\\a_{n}=266666.67(-\frac{1}{2})^{n}+233333.33[/tex]
Hope this helps!