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Using generating functions, solve each LHRRWCC.$$a_{n}=a_{n-1}+a_{n-2}, a_{0}=1, a_{1}=2$$...

Question

Using generating functions, solve each LHRRWCC.$$a_{n}=a_{n-1}+a_{n-2}, a_{0}=1, a_{1}=2$$

Using generating functions, solve each LHRRWCC. $$ a_{n}=a_{n-1}+a_{n-2}, a_{0}=1, a_{1}=2 $$



Answers

Let $\left\{a_{n}\right\}$ be the sequence defined recursively by
$$
a_{0}=0, \quad a_{n+1}=\sqrt{2+a_{n}}
$$
Thus, $a_{1}=\sqrt{2}, \quad a_{2}=\sqrt{2+\sqrt{2}}, \quad a_{3}=\sqrt{2+\sqrt{2+\sqrt{2}}}, \ldots$
$$
\begin{array}{l}{\text { (a) Show that if } a_{n}<2, \text { then } a_{n+1}<2 . \text { Conclude by induction that }} \\ {a_{n}<2 \text { for all } n .} \\ {\text { (b) Show that if } a_{n}<2, \text { then } a_{n} \leq a_{n+1} . \text { Conclude by induction that }} \\ {\left\{a_{n}\right\} \text { is increasing. }} \\ {\text { (c) Use (a) and (b) to conclude that } L=\lim _{n \rightarrow \infty} a_{n} \text { exists. Then compute }} \\ {L \text { by showing that } L=\sqrt{2+L}}\end{array}
$$

Hello, everyone. My name, Mrs Smith. And I'm going to solve a problem from chapter 11. Problem number 50. We are given with the first term. Justin and Eric. Also formula Let us find other domes in the sequence. Hey, two equals two times human, which is 20. It recalls two times a two. It is 40 Hey for equals, two times 83 It is 80. As you can see, there isn't a constant multiplication between these terms hands. This is not an automatic siris. No, we have to take for the tumor. Take cities. As you can see, 20 can be to turn us due to the part off One into 10 on 40 can be written as two part two in tow. 10. Similarly, 80 can bitterness to part three in tow. Then hence our explicit formula would be e off and equals to fire in minus one into then

Hello, everyone, my name, A system and I'm going to sort of problem from Chapter 11 Problem number 54. Use each record fuel formula to write an explicit formula for the sequence we're going with the first term as minus two on Erica's a formula Let's right, o other terms a two equals on by two times even witches on by two times minus two, which is minus one a three equals or my do times What a to which is owned by two times minus one, which is minus one by two. A four equals one by two times a tree, which is one by two times my list one by two, it is minus one by four. As you can see, there isn't a constant. The office is getting ready to play to get a new. So as this is not an automatic progression, we have to check for geometric progression. So this ah can be return as group are zero, this is DuPont minus one. This is two parts minus go. So this is to about minus one to a prom minus two group on ministry and do minus two so hands we can read the explicit formula ass to par one minus and times minus two. So if an equals to bar one minus end times minus two. So let us take the formula. Wants be subdued and equals one a of one equal Stupar one minus one times minus two, but is equal. So this comes to the part of zero. Minister has this is minus two and equals three. A three equals to fire one minus three times minus two, which is group are minus two times minus two inches. One by four times minus two. It is equals two minus one vato. Hence this is the

Okay, so we're given this recursive formula and were asked to write the explicit formula for it. So whenever we're looking for an explicit formula, we want to find the relationship between n and A. M were given and the first term in the sequence a one equals one and find the second we could just go along this rule here, whatever the previous term is plus four. So one plus for us five five plus four is 99 plus four who was 13 and those of the third and fourth terms. Now we're looking for relationship between an and A N, and we're trying to figure out a connection a way that we could get this number by manipulating this number. Well, all of these are very close to multiples of four. Right? Like five is just one off from 49 isjust one off from eight and 13 is just one off from 12. Which makes sense, since we're adding for every time starting at one. So if one is our starting point and for and numbers down the line, we've added four and minus one times, so n minus one times where adding four toe one right, because we started with one and every time we moved down here we've added for another time. So when n equals two, we've added for once when n equals three, we've added for twice when articles four, we've added for three times, but our starting value was one. So this looks this could be our formula because it's based off of end. And it would give us this number, right? We look, I say we start with one one minus one is zero zero times for zero and then we're just left with one, which gives us they're starting for two two minus one is 11 times four is four and four plus one is five three minus one is 22 times four is eight plus one. It gives us nine. So this fits. However, it's kind of messy looking, so let's try and simplify it. Maybe distribute this four into this and minus ones we get for and minus four instead. And then we can simplify to or an negative four plus one. Because this negative three, which still fits all were sequence try one times four is four minus three is one two times four is eight minus three is five. Three times four is 12 minus three is nine. So it works just fine, and this is our explicit formula for the sequence.

We're trying to find out the first five numbers in this sequence. Now, this is a recursive sequence, and that means the number that comes next depends on the numbers that came before it. Specifically, we're gonna be summing the two previous amounts. Now they gave us our first two. So a one is one and a two is two. Now, for a three, we have to go to the one right before it and minus one, which is two. And we're adding to that, the one that came before that which in this case is one. So our third term is three because we added two and one together. Now, for our fourth term, we're gonna be adding the one before it, which is three and the one that became for that, which is, too. Which means that our fourth term is five. And for 1/5 term, we're gonna be adding five, which was a four and three, which was a three now. Five plus three is eight. So the first five numbers are 1235 and eight


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