Proof involving linear algebra

spoc21
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Homework Statement


Hi, I'm supposed to solve the following question using proof by induction, and am very confused with it. It would be greatly appreciated if someone could help me solve this problem:

Let an = 2 and an+1\frac{4a_n -3}{a_n} for n >=1. Show that 1\leq a_n \leq a_(n+1)\leq3 for all n \geq1
please note that _ = subscript I am very confused with this problem, and would appreciate any help.Thanks!
 
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What you posted doesn't make sense. Check to make sure that it's written correctly.
 
spoc21 said:

Homework Statement


Hi, I'm supposed to solve the following question using proof by induction, and am very confused with it. It would be greatly appreciated if someone could help me solve this problem:

Let an = 2 and an+1\frac{4a_n -3}{a_n} for n >=1. Show that 1\leq a_n \leq a_(n+1)\leq3 for all n \geq1
please note that _ = subscript


I am very confused with this problem, and would appreciate any help.


Thanks!

I have myself solved something simular not so long a ago I think it suppose to say

Let a_n = 2

Let a_{n+1} = \frac{4 {a_n}-3}{a_n} for n \geq 1

and then

"Show that 1\leq {a_n} \leq {a_{n+1}}\leq 3 for all n \geq 1"

I could be wrong but this setup makes me think about a socalled recurrence relation..
But you are right it looks like something is missing.
 
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Have you tried anything yet? Induction has some very clear steps. First, show it is true for n=1. Then assume n=k is true and see what happens with k+1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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