Exercise 5 (5 points) — Candidates who have not followed the speciality courseLet $( u _ { n } )$ be the sequence defined by $u _ { 0 } = 3 , u _ { 1 } = 6$ and, for all natural integer $n$:
$$u _ { n + 2 } = \frac { 5 } { 4 } u _ { n + 1 } - \frac { 1 } { 4 } u _ { n } .$$
The purpose of this exercise is to study the possible limit of the sequence $( u _ { n } )$.
Part A:
We wish to calculate the values of the first terms of the sequence $( u _ { n } )$ using a spreadsheet. We have reproduced below part of a spreadsheet, where the values of $u _ { 0 }$ and $u _ { 1 }$ appear.
| A | B |
| 1 | $n$ | $u _ { n }$ |
| 2 | 0 | 3 |
| 3 | 1 | 6 |
| 4 | 2 | |
| 5 | 3 | |
| 6 | 4 | |
| 7 | 5 | |
- Give a formula which, entered in cell B4, then copied downwards, allows obtaining values of the sequence $( u _ { n } )$ in column B.
- Copy and complete the table above. Approximate values to $10 ^ { - 3 }$ of $u _ { n }$ will be given for $n$ ranging from 2 to 5.
- What can be conjectured about the convergence of the sequence $( u _ { n } )$?
Part B: Study of the sequence
We consider the sequences $( v _ { n } )$ and $( w _ { n } )$ defined for all natural integer $n$ by:
$$v _ { n } = u _ { n + 1 } - \frac { 1 } { 4 } u _ { n } \quad \text { and } \quad w _ { n } = u _ { n } - 7 .$$
- a. Prove that $( v _ { n } )$ is a constant sequence. b. Deduce that, for all natural integer $n , u _ { n + 1 } = \frac { 1 } { 4 } u _ { n } + \frac { 21 } { 4 }$.
- a. Using the result of question 1. b., show by induction that, for all natural integer $n , u _ { n } < u _ { n + 1 } < 15$. b. Deduce that the sequence $( u _ { n } )$ is convergent.
- a. Prove that $( w _ { n } )$ is a geometric sequence and specify its first term and common ratio. b. Deduce that, for all natural integer $n , u _ { n } = 7 - \left( \frac { 1 } { 4 } \right) ^ { n - 1 }$. c. Calculate the limit of the sequence $( u _ { n } )$.