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Sequences revision

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Tutor: Labib

Summary

Sequences revision

In a nutshell

General sequences are a list of numbers which can either be described by a term to term rule or position to term rule.  These sequences have various patterns and may or may not have limits. 



Defining a sequence

The term to term rule gives a term in a general sequence and the calculation to go to the next term. The position to term rule uses an equation to work out a number in a sequence if the position in the sequence is known. 


Example 1

Given that  u1=10u_1 = 10 and un+1=2un+3u_{n+1} = 2u_n + 3, what is the 3rd3^{rd} term in the sequence?


The number nn indicates the position in the sequence. So u1u_1 is the first term in the sequence. To find u2u_2 (the second term), substitute n=1n=1 into the equation of the sequence:

u1+1=2u1+3u2=2(10)+3\begin{aligned} u_{1+1} &= 2u_1 +3 \\ u_2 &= 2(10) + 3 \end{aligned}​​

u2=23u_2 = 23


Substitute n=2n=2 into the equation of the sequence to find the third term:

u2+1=2u2+3u3=2(23)+3\begin{aligned} u_{2+1} &= 2u_2 +3 \\ u_3 &= 2(23) + 3 \end{aligned}​​

u3=49\underline{u_3 = 49}​​


Example 2

What is the 6th6^{th} term in a sequence with the formula u2n=4n+52u_{2n} =\dfrac{ 4n +5}{2}?


Find the value of nn which will give the 6th6^{th} term:

u2n=u62n=6n=3\begin{aligned} u_{2n}&=u_6\\2n &= 6 \\ n&=3 \end{aligned}​​


Substitute n=3n=3 into the formula:

u2(3)=4(3)+52u_{2(3)} = \dfrac {4(3) + 5 } {2}


u6=172=8.5\underline { u_6 = \dfrac {17}{2}= 8.5 }​​



Increasing, decreasing or periodic sequences

Sequences may increase such that each term is always larger than the previous one, they may decrease such that each term is smaller than the previous one or they may be periodic such that the sequence eventually repeats itself.


Example 3

Find the first 44 terms of un+1=1unu_{n+1} = \dfrac{1}{u_n} given that u1=5u_1 = 5. Describe the behaviour of this sequence. 


Calculate u2, u3, u4u_2, \ u_3,\ u_4:

u2=15u3=115=5u4=15\begin{aligned} u_2 &= \dfrac{1}{5} \\ \\ u_3 &= \dfrac{1}{\frac15} = 5 \\ \\ u_4 &=\dfrac {1}{5} \end{aligned}​​


Describe the behaviour of the sequence:

The sequence is periodic with a periodicity of 2\underline2.


Note: Periodicity means how many terms there are that the sequence goes through before repeating.



Limits

A given general sequence may be convergent or divergent. A convergent sequence has a limit such that as nn \rightarrow \infin ("nn tends to infinity") the values of unu_n  and un+1u_{n+1} will be the same. A divergent sequence has no limit, so as nn \rightarrow \infin​ the values of unu_n ​ and un+1u_{n+1} will continue to be different. 


Example 4

A convergent sequence is defined as un+1=35un+4u_{n+1} = \dfrac35 u_n +4. What is the limit, LL?


Consider nn \rightarrow \infinYou know that the sequence is convergent so there exists a limit LL such that

un+1=un=Lu_{n+1} = u_n = L​​


Using the sequence equation, solve for LL:

L=35(L)+4L = \dfrac 35 (L) + 4

​​

25(L)=4\dfrac 25 (L) = 4

​​

2L=20L=102L = 20 \\ \underline{L = 10}​​



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