A2 June 2019 Q5

EdexcelCurrent spec11 marksRecurrence Relations

5. An increasing sequence \(\{u_n\}\) for \(n \in \mathbb{N}\) is such that the difference between the \(n\)th term of \(\{u_n\}\) and the mean of the previous two terms of \(\{u_n\}\) is always 6

(a) Show that, for \(n \geqslant 3\)\[2u_n - u_{n-1} - u_{n-2} = 12\] (2)

Given that \(u_1 = 2\) and \(u_2 = 8\)

(b) find the solution of this second order recurrence relation to obtain an expression for \(u_n\) in terms of \(n\). (7)

(c) Show that as \(n \to \infty\), \(u_n \to kn\) where \(k\) is a constant to be determined. You must give reasons for your answer. (2)

Mark scheme (a)

Scheme	Marks	AO
\(u_n - \tfrac{1}{2}(u_{n-1} + u_{n-2}) = 6 \Rightarrow 2u_n - (u_{n-1} + u_{n-2}) = 12\)	1M1	2.1
\(2u_n - u_{n-1} - u_{n-2} = 12\)	1A1	2.2a
	(2)

Notes

1M1: attempt to write given information in terms of a recurrence equation (allow sign errors and errors in notation)

1A1*: CAO (NB equation is provided in the question so must see a correct unsimplified recurrence relation which is simplified with no errors to the required form for this mark).

Mark scheme (b)

Scheme	Marks	AO
aux equation \(2m^2 - m - 1 = 0 \Rightarrow m = 1, m = -\tfrac{1}{2}\)	1B1	2.1
\(u_n = A + B\left(-\tfrac{1}{2}\right)^n\)	2B1	1.1b
particular solution try \(u_n = \lambda n\) \(\therefore 2\lambda n - \lambda n + \lambda - \lambda n + 2\lambda = 12 \Rightarrow \lambda [= 4]\)	1M1	1.1b
\(u_n = A + B\left(-\tfrac{1}{2}\right)^n + 4n\)	1A1	1.1b
\(u_1 = 2 \Rightarrow 2A - B = -4\)	2M1	1.1b
\(u_2 = 8 \Rightarrow 4A + B = 0\)	3M1	1.1b
\(A = -\tfrac{2}{3}, B = \tfrac{8}{3} \Rightarrow u_n = -\tfrac{2}{3} + \tfrac{8}{3}\left(-\tfrac{1}{2}\right)^n + 4n\)	2A1	2.2a
	(7)

Notes

1B1: CAO for auxiliary equation and corresponding solutions (this mark can be implied by correct complementary function)

2B1: complementary function CAO. Condone lack of ‘\(u_n =\)’.

1M1: substitutes \(u_n = \lambda n\) into their 2nd-order recurrence relation and solves to obtain \(\lambda = \ldots\)

Note: \(2\lambda n - \lambda(n - 1) - \lambda(n - 2) = 12\) o.e. \(\Rightarrow \lambda = \cdots\) can earn this mark

Note: Do not condone sign errors or errors in coefficients e.g.
\(2\lambda n - \lambda(n + 1) - \lambda(n - 2) = 12\) is M0
\(2\lambda n - 2\lambda(n - 1) - \lambda(n - 2) = 12)\) is M0

1A1: Correct general solution. Condone lack of ‘\(u_n =\)’.

2M1: Forms one equation in \(A\) and \(B\). General Solution must be of the form.
\(u_n = A + B\left(-\tfrac{1}{2}\right)^n + \mu n\) where \(\mu \neq 0\)

3dM1: Forms a second equation in \(A\) and \(B\). Dependent on the previous method mark.

2A1: Particular solution CAO. Do not condone lack of ‘\(u_n = \ldots\)’

Mark scheme (c)

Scheme	Marks	AO
\(u_n \to 4n \qquad (k = 4)\) As \(n \to \infty\), \(\left(-\tfrac{1}{2}\right)^n \to 0\) and \(4n\) is considerably greater than \(-\tfrac{2}{3}\)	1M1 1A1ft	2.1 1.1b
	(2)
	(11 marks)

Notes

M1: Obtains correct limit (ft their particular solution which must be of the correct form i.e. \(u_n = A + B\left(-\tfrac{1}{2}\right)^n + \mu n\) where \(\mu \neq 0\))

A1ft: Provides correct reasoning including comments relating to both of the following:

\(n \to \infty, \left(-\tfrac{1}{2}\right)^n \to 0\)
‘\(-\tfrac{2}{3}\)’ becomes negligible (compared to \(4n\) as \(n \to \infty\))

Note: Condone ‘\(-\tfrac{2}{3}\) becomes insignificant’