C1 June 2013 (R) Q6

EdexcelOld spec9 marksSequences & Series

6. A sequence \(x_1, x_2, x_3, \ldots\) is defined by\[\begin{aligned} x_1 &= 1 \\ x_{n+1} &= (x_n)^2 - kx_n, \quad n \geqslant 1 \end{aligned}\]where \(k\) is a constant, \(k \neq 0\)

(a) Find an expression for \(x_2\) in terms of \(k\). (1)

(b) Show that \(x_3 = 1 - 3k + 2k^2\) (2)

Given also that \(x_3 = 1\),

(d) Hence find the value of \(\displaystyle\sum_{n=1}^{100} x_n\) (3)

Scheme	Marks
\(x_2 = 1 - k\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(x_3 = (1 - k)^2 - k(1 - k)\)	M1
\(= 1 - 3k + 2k^2\,*\)	A1*
	(2)

Notes

M1: Attempt to substitute their \(x_2\) into \(x_3 = (x_2)^2 - kx_2\) with their \(x_2\) in terms of \(k\).

A1*: Answer given

Mark scheme (c)

Scheme	Marks
\(1 - 3k + 2k^2 = 1\)	M1
\(\left(2k^2 - 3k = 0\right)\)
\(k(2k - 3) = 0 \Rightarrow k = ..\)	dM1
\(k = \dfrac{3}{2}\)	A1
	(3)

Notes

M1: Setting \(1 - 3k + 2k^2 = 1\)

dM1: Solving their quadratic to obtain a value for \(k\). Dependent on the previous M1.

A1: Cao and cso (ignore any reference to \(k = 0\))

Mark scheme (d)

Scheme	Marks
\(\displaystyle\sum_{n=1}^{100} x_n = 1 + \left(-\frac{1}{2}\right) + 1 + \ldots\ldots\) Or \(= 1 + (1 - \text{‘}k\text{’}) + 1 + \ldots\ldots\)	M1
\(50\times\dfrac{1}{2}\) or \(50\times 1 - 50\times\dfrac{1}{2}\) or \(\dfrac{1}{2}\times 100\times\left(1 - \dfrac{1}{2}\right)\)	M1
\(= 25\)	A1
Note that the use of \(\dfrac{1}{2}n(a + l)\) is acceptable here but \(\dfrac{1}{2}n(2a + (n - 1)d)\) is not.
Allow correct answer only
	(3)
	(9 marks)

Notes

M1: Writing out at least 3 terms with the third term equal to the first term. Allow in terms of \(k\) as well as numerical values.
Evidence that the sequence is oscillating between 1 and \(1 - k\).
This may be implied by a correct sum.

M1: An attempt to combine the terms correctly. Can be in terms of \(k\) here e.g \(100 - 50k\)

A1: Allow an equivalent fraction, e.g. 50/2 or 100/4

(corrected from the printed mark scheme: the third expression is printed as \(\tfrac{1}{2}\times 50\times(1 - \tfrac{1}{2})\), which equals 12.5; the \(\tfrac{1}{2}n(a + l)\) form needs \(n = 100\).)