C1 January 2012 Q4

EdexcelOld spec6 marksSequences & Series

4. A sequence \(x_1, x_2, x_3, \ldots\) is defined by\[\begin{aligned}x_1 &= 1\\ x_{n+1} &= ax_n + 5, \qquad n \geqslant 1\end{aligned}\]where \(a\) is a constant.

(a) Write down an expression for \(x_2\) in terms of \(a\). (1)

(b) Show that \(x_3 = a^2 + 5a + 5\) (2)

Given that \(x_3 = 41\)

Scheme	Marks
\(\left(x_2 =\right)\ a + 5\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(\left(x_3\right) = a\text{"}(a + 5)\text{"} + 5\)	M1
\(= a^2 + 5a + 5 \qquad (*)\)	A1cso
	(2)

Notes

M1: must see \(a(\text{ their } x_2) + 5\)

A1cso: must have seen \(a(\,a[1] + 5) + 5\) (etc or better) Must have both brackets (....) and no incorrect working seen

Mark scheme (c)

Scheme	Marks
\(41 = a^2 + 5a + 5 \ \Rightarrow a^2 + 5a - 36\,(= 0)\) or \(36 = a^2 + 5a\)	M1
\((a + 9)(a - 4) = 0\)	M1
\(a = 4\) or \(-9\)	A1
	(3)
	(6 marks)

Notes

1^st M1: for forming a suitable equation using \(x_3\) and 41 and an attempt to collect like terms and reduce to 3TQ (o.e.). Allow one error in sign. Accept for example \(a^2 + 5a + 46\,(= 0)\)
If completing the square should get to \(\left(a \pm \tfrac{5}{2}\right)^2 = 36 + \tfrac{25}{4}\)

2^nd M1: Attempting to solve their relevant 3TQ (see General Principles)

A1: for both 4 and \(-9\) seen. If \(a = 4\) and \(-9\) is followed by \(-9 < a < 4\) apply ISW.
No working or trial and improvement leading to both answers scores 3/3 but no marks for only one answer.

Allow use of other letters instead of \(a\)