Edexcel IAL WMA12/01/P2/Jan 2021/Q8 (Series, Recurrence Relationships)
A sequence a1 , a2, a3 , … is defined by
an+1 = 2an + 32 - 7
a1 = p - 3
where p is a constant.
(1)
Given that an
∑n=13an=p+15
(6)
SOLUTION
a- To find a2, simply use the nth term formula given in the question.
an+1 = 2an + 32 - 7
We will put n=1, so that on the left hand side of the equation, we get an+1=a1+1=a2
a1+1 = 2a1 + 32 - 7
a2=2 a1+32-7
a2=2p-3+32-7
a2=2p2-7
b-
it is given that the sum of the first 3 terms of the series, which expression is given as ∑n=13an, is equal to p+15 . So, we are first going to find the sum of the first three terms by epanding the summation expression and then will equate it to p+15.
a1+a2+a3=p+15
Where for a3 we need to put n=2.
an+1 = 2an + 32 - 7
a2+1 = 2a2 + 32 - 7
a3=2 a2+32-7
Substituting the value of a2 in the above expression.
a3=22p2-7+32-7
a3=22p2-42-7
a3=24p4-16p2+16-7
a3=8p4-32p2+32-7
a3=8p4-32p2+25
Now, summing all the first 3 terms of the expression and equating it to p+15.
a1+a2+a3=p+15
p-3+2p2-7+8p4-32p2+25=p+15
8p4-30p2+p+15=p+15
8p4-30p2=0
8p4-30p2=0
4p4-15p2=0
p24p2-15=0
p2=0 4p2=15
p=0 p2=154
p= ±15 2
Substituting the both values of p to find the possible value of a2.
a2=2p2-7
When, p=0
a2=202-7
a2= -7
When, p= ±152
a2=2± 1522-7
a2=2154-7=7.5 -7
a2=0.5
Hence, the possible values of
a2=-7 & 0.5
