The discussion revolves around a recurrence relation defined as a_{n+2}=3a_{n+1}-2a_n with initial conditions a_1=1 and a_2=1. Participants are tasked with finding the next five terms of the sequence.
The discussion includes attempts to clarify the recurrence relation and its implications. Some participants have provided guidance on how to compute the next term, while others are exploring the nature of the sequence and whether it leads to a constant value.
There is a noted uncertainty about the implications of the recurrence relation, particularly regarding the values of the terms beyond the initial conditions. Participants are also navigating the challenge of understanding the notation and its application in this context.
The terms in the sequence are {a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ..., a<sub>n</sub>, a<sub>n+1</sub>, a<sub>n+2</sub>, ...}.<br /> <br /> The first formula says that to get the (n + 2)nd term in the sequence you need the preceding two terms, the (n + 1)st term and the nth term. <blockquote data-attributes="" data-quote="cdotter" data-source="post: 2668920" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> cdotter said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Could someone do the next few terms so I can see how it's done? </div> </div> </blockquote>Well, no, but maybe you can do them. You have a<sub>1</sub> = 1 and a<sub>2</sub> = 1. Use the first formula to get a<sub>3</sub>. Then when you have a<sub>3</sub>, use the formula again to find a<sub>4</sub>, and so on for as many terms as you need.cdotter said:Homework Statement
a_{n+2}=3a_{n+1}-2a_n[/itex]<br /> a_1=1, a_2=1<br /> Find the next 5 terms.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> I don't really understand the "a sub n" notation.<br />
The formula iscdotter said:a_{1+2}=3a_{1+1} + 2a_1
a_3 = 3(1)-2(1) = 1?
For this sequence, yes.cdotter said:So any possible a_n always equals 1?
Mark44 said:For this sequence, yes.
A simpler and nonrecursive definition would be an = 1 for n = 1, 2, 3, ...