Edexcel IAL WMA12/01/P2/June 2019/Q1 (Recurrence Relationships)
Answer all questions. Write your answers in the spaces provided
A sequence a1, a2, a3, …. is defined by
an+1=4-an
a1=3
Find the value of
(2)
(b) ∑n=12000(2an-1)
(2)
SOLUTION
a- i- For a2, put n=1
an+1=4-an
a1+1=4-a1
Substituting a1=3
a2=4-3
=4-3
a2=1
a-ii- Before we find a107, lets see whats the patterm looks like.
a2+1=4-a2
a3=4-1
a3=3
For a4,
a3+1=4-a3
a4=4-3
a4=1
Thus, this gives us alternating pattern: 3, 1, 3, 1, ….
Each odd term is 3 and even term is 1.
Now, the a107 is an odd term so it must be 3.
Hence,
a107=3
b- Lets split the given summation equation.
∑n=12000(2an-1)=2∑n=1200an- ∑n=12001
For ∑n=12002an, an is going to be 3+1+3+1+. . because it also follows the same pattern (all odd terms are 3 and all even terms will be 1).
∑n=12002an=23+1+3+1+. . ..
Whereas, ∑n=12001, we are going to sum 200 terms where each term is 1.
∑n=12001=200(1+1+1+ .. . 1)
Hence,
∑n=12000(2an-1)=2∑n=1200an- ∑n=12001
∑n=12000(2an-1)=23+1+3+1+. . ..-2001+1+1+ .. . 1
∑n=12000(2an-1)=2 100 x 3+100 x 1-200
∑n=12000(2an-1)=2 400-200
∑n=12000(2an-1)=600
