Standard +0.8 This is a Further Maths question requiring students to verify a particular solution form for a recurrence relation by determining constants through substitution and algebraic manipulation. While the verification process is methodical rather than requiring deep insight, it involves factorial notation, pattern recognition, and careful algebraic work that goes beyond standard A-level content, placing it moderately above average difficulty.
2 A sequence is defined by the recurrence relation \(t _ { n + 1 } = \frac { t _ { n } } { n + 3 }\) for \(n \geqslant 1\), with \(t _ { 1 } = 8\).
Verify that the particular solution to the recurrence relation is given by \(t _ { n } = \frac { a } { ( n + b ) ! }\) where \(a\) and \(b\) are constants whose values are to be determined.
Question 2:
2 | a
t = =8 soi
1 (1+b)!
t a a
t = n ⇒ =
n+1 n+3 (n+1+b)! (n+3)(n+b)!
so we need n + 3 = n + 1 + b
=> b = 2
48
and a = 8×3! = 48 t =
n (n+2)!
� � | B1
M1
M1
A1
A1
[5] | 3.1a
3.1a
1.1
1.1
3.2a | Using the initial condition to
obtain an equation in a and b
Substituting solution formula into
recurrence relation
Cancelling a and (n + b)!
2 | Alternative Method:
a
t = =8
1 (1+b)!
𝑎𝑎
𝑡𝑡2=(2+𝑏𝑏)(1+𝑏𝑏)!=2
Solving to give a = 48, b = 2
48
𝑡𝑡 𝑛𝑛+1 =
(𝑛𝑛+2)!(𝑛𝑛+3)
48
𝑡𝑡 𝑛𝑛+1 =
(𝑛𝑛+3)! | M1
A1A1
M1
A1
[5] | 3.1a
3.1a
1.1
3.2a
Attempting to find two terms eg t
1
and t or recognising a general
2
pattern
Using to verify
𝑡𝑡𝑛𝑛
solution
𝑛𝑛+1
𝑡𝑡 =𝑛𝑛+3
Completion
Alternative Method:
a
t = =8
1 (1+b)!
𝑎𝑎
𝑡𝑡2=(2+𝑏𝑏)(1+𝑏𝑏)!=2
Solving to give a = 48, b = 2
48
𝑡𝑡 𝑛𝑛+1 =
(𝑛𝑛+2)!(𝑛𝑛+3)
48
𝑡𝑡 𝑛𝑛+1 =
(𝑛𝑛+3)!
M1
A1A1
M1
A1
[5]
3.1a
3.1a
1.1
3.2a
2 A sequence is defined by the recurrence relation $t _ { n + 1 } = \frac { t _ { n } } { n + 3 }$ for $n \geqslant 1$, with $t _ { 1 } = 8$.\\
Verify that the particular solution to the recurrence relation is given by $t _ { n } = \frac { a } { ( n + b ) ! }$ where $a$ and $b$ are constants whose values are to be determined.
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2020 Q2 [5]}}