Phillip R. answered • 08/16/14

Top Notch Math and Science Tutoring from Brown Univ Grad

I believe the solution is a summation for n ≥ 0 rather than the values of each term.

a

_{n+1}- a_{n-1}= 2a_{n }and be rearranged toa

_{n+1}= 2a_{n}+ a_{n-1}The subscripts can be changed to n, n-1, and n-2 so we have an expression for a_{n}in terms of two previous values in the sequence.Starting with a

_{n}= 2a_{n-1}+ a_{n-2}you can make successive substitutions for a_{n-1}, then a_{n-2}, and so on until you get to a_{0.}a

_{4}has the value 12a_{1}+ 5a_{0}a

_{6}has the value 70a_{1}+ 29a_{0}What I am unable to get is a similar expression for the general term a

_{n}Maybe we are supposed to stop at a

_{2}given a_{0}and a_{1}are defined initial values.I am hoping someone can clarify this.