Standard +0.3 This is a standard convergence problem requiring students to set u_{n+1} = u_n = α and solve the resulting quadratic equation α = 12/(1+α). While it involves a recurrence relation and limit concept from Further Maths, the solution method is routine and requires only basic algebraic manipulation with no novel insight or extended reasoning.
1 The sequence \(\left\{ u _ { n } \right\}\) is defined by \(u _ { 1 } = 2\) and \(u _ { n + 1 } = \frac { 12 } { 1 + u _ { n } }\) for \(n \geq 1\).
Given that the sequence converges, with limit \(\alpha\), determine the value of \(\alpha\).
Set u (cid:32)u (cid:32)(cid:68)(cid:159) (cid:68)(cid:32)
n (cid:14) 1 n 1(cid:14)(cid:68)
(cid:68) = 3 and/or (cid:68) = – 4
Answer
Marks
Since (cid:68) > 0 (all terms are positive), (cid:68) = 3
M1
A1
E1
Answer
Marks
[3]
1.1
1.1
Answer
Marks
2.2a
Use of the limit in the given
reccurence relation
BC Solving for (cid:68)
Explicitly rejecting the negative
solution or justifying why there is
Answer
Marks
only one positive solution
OR BC
M1 “2” ENTER and iterating with 12
/ (1 + ANS)
A1 (cid:111) (cid:68) = 3
Question 1:
1 | 12
Set u (cid:32)u (cid:32)(cid:68)(cid:159) (cid:68)(cid:32)
n (cid:14) 1 n 1(cid:14)(cid:68)
(cid:68) = 3 and/or (cid:68) = – 4
Since (cid:68) > 0 (all terms are positive), (cid:68) = 3 | M1
A1
E1
[3] | 1.1
1.1
2.2a | Use of the limit in the given
reccurence relation
BC Solving for (cid:68)
Explicitly rejecting the negative
solution or justifying why there is
only one positive solution | OR BC
M1 “2” ENTER and iterating with 12
/ (1 + ANS)
A1 (cid:111) (cid:68) = 3
1 The sequence $\left\{ u _ { n } \right\}$ is defined by $u _ { 1 } = 2$ and $u _ { n + 1 } = \frac { 12 } { 1 + u _ { n } }$ for $n \geq 1$.\\
Given that the sequence converges, with limit $\alpha$, determine the value of $\alpha$.
\hfill \mbox{\textit{OCR Further Additional Pure AS Q1 [3]}}