OCR Further Additional Pure AS 2019 June — Question 2 4 marks

Exam BoardOCR
ModuleFurther Additional Pure AS (Further Additional Pure AS)
Year2019
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and Series
TypeConvergence and Limits of Sequences
DifficultyChallenging +1.2 This is a Further Maths question on recursive sequences and limits, requiring students to set L = √L + 4/√L and solve the resulting equation. While it involves Further Maths content (convergence of sequences), the technique is standard: assume convergence to limit L, substitute into the recurrence relation, and solve algebraically. Part (b) is a straightforward extension using the same method with a parameter. The algebraic manipulation is routine for Further Maths students, making this moderately above average difficulty but not requiring novel insight.
Spec8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states

2
  1. The convergent sequence \(\left\{ \mathrm { a } _ { \mathrm { n } } \right\}\) is defined by \(a _ { 0 } = 1\) and \(\mathrm { a } _ { \mathrm { n } + 1 } = \sqrt { \mathrm { a } _ { \mathrm { n } } } + \frac { 4 } { \sqrt { \mathrm { a } _ { \mathrm { n } } } }\) for \(n \geqslant 0\). Calculate the limit of the sequence.
  2. The convergent sequence \(\left\{ \mathrm { b } _ { \mathrm { n } } \right\}\) is defined by \(\mathrm { b } _ { 0 } = 1\) and \(\mathrm { b } _ { \mathrm { n } + 1 } = \sqrt { \mathrm { b } _ { \mathrm { n } } } + \frac { \mathrm { k } } { \sqrt { \mathrm { b } _ { \mathrm { n } } } }\) for \(n \geqslant 0\), where \(k\) is a constant. Determine the value of \(k\) for which the limit of the sequence is 9 .

Question 2:
AnswerMarks Guidance
2(a) Limit = 4
[1]1.1 BC
(b)Setting b = b = 9 throughout
n + 1 n
k
Solving 9 9 
9
AnswerMarks
k = 18M1
M1
AnswerMarks
A13.1a
1.1a
AnswerMarks
1.1soi
BC or by inspection
Alternative method
k
Search method for b = b  with various k’s
n + 1 n
b
AnswerMarks
nM1
Evidence of systematic approach (e.g. k = 9  6.11…, k = 20  9.56…)M1
k = 18A1
[3]
Question 2:
2 | (a) | Limit = 4 | B1
[1] | 1.1 | BC
(b) | Setting b = b = 9 throughout
n + 1 n
k
Solving 9 9 
9
k = 18 | M1
M1
A1 | 3.1a
1.1a
1.1 | soi
BC or by inspection
Alternative method
k
Search method for b = b  with various k’s
n + 1 n
b
n | M1
Evidence of systematic approach (e.g. k = 9  6.11…, k = 20  9.56…) | M1
k = 18 | A1
[3]
2
\begin{enumerate}[label=(\alph*)]
\item The convergent sequence $\left\{ \mathrm { a } _ { \mathrm { n } } \right\}$ is defined by $a _ { 0 } = 1$ and $\mathrm { a } _ { \mathrm { n } + 1 } = \sqrt { \mathrm { a } _ { \mathrm { n } } } + \frac { 4 } { \sqrt { \mathrm { a } _ { \mathrm { n } } } }$ for $n \geqslant 0$. Calculate the limit of the sequence.
\item The convergent sequence $\left\{ \mathrm { b } _ { \mathrm { n } } \right\}$ is defined by $\mathrm { b } _ { 0 } = 1$ and $\mathrm { b } _ { \mathrm { n } + 1 } = \sqrt { \mathrm { b } _ { \mathrm { n } } } + \frac { \mathrm { k } } { \sqrt { \mathrm { b } _ { \mathrm { n } } } }$ for $n \geqslant 0$, where $k$ is a constant.

Determine the value of $k$ for which the limit of the sequence is 9 .
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure AS 2019 Q2 [4]}}