C1 June 2015 Q4

EdexcelOld spec8 marksSequences & Series

(i) A sequence \(U_1, U_2, U_3, \ldots\) is defined by\[\begin{gathered} U_{n+2} = 2U_{n+1} - U_n, \quad n \geqslant 1 \\ U_1 = 4 \text{ and } U_2 = 4 \end{gathered}\]Find the value of

(a) \(U_3\) (1)

(b) \(\displaystyle\sum_{n=1}^{20} U_n\) (2)

(ii) Another sequence \(V_1, V_2, V_3, \ldots\) is defined by\[\begin{gathered} V_{n+2} = 2V_{n+1} - V_n, \quad n \geqslant 1 \\ V_1 = k \text{ and } V_2 = 2k, \text{ where } k \text{ is a constant} \end{gathered}\]

(a) Find \(V_3\) and \(V_4\) in terms of \(k\). (2)

Given that \(\displaystyle\sum_{n=1}^{5} V_n = 165\),

(b) find the value of \(k\). (3)

Scheme	Marks
\(U_3 = 4\)	B1
	(1)

Mark scheme (i)(b)

Scheme	Marks
\(\displaystyle\sum_{n=1}^{n=20} U_n = 4 + 4 + 4 \ldots\ldots\ldots + 4\) or \(20\times 4\)	M1
\(= 80\)	A1
	(2)

Notes

M1: For realising that all 20 terms are 4 and that the sum is required. Possible ways are 4+4+4…..+4 or \(20\times 4\) or \(\dfrac{1}{2}\times 20\left(2\times 4 + 19\times 0\right)\) or \(\dfrac{1}{2}\times 20\left(4 + 4\right)\)
(Use of a correct sum formula with \(n = 20\), \(a = 4\) and \(d = 0\) or \(n = 20\), \(a = 4\) and \(l = 4\))

A1: cao

Correct answer with no working scores M1A1

Scheme	Marks
\(V_3 = 3k,\ \ V_4 = 4k\)	B1, B1
	(2)

Mark scheme (ii)(b)

Scheme	Marks
\(\displaystyle\sum_{n=1}^{n=5} V_n = k + 2k + 3k + 4k + 5k = 165\) or \(\dfrac{1}{2}\times 5\left(2\times k + 4\times k\right) = 165\) or \(\dfrac{1}{2}\times 5\left(k + 5k\right) = 165\)	M1
\(15k = 165 \Rightarrow k = ..\)	M1
\(k = 11\)	A1
	(3)
	(8 marks)

Notes

M1: Attempts \(V_5\), adds their \(V_1, V_2, V_3, V_4, V_5\) AND sets equal to 165
or
Use of a correct sum formula with \(a = k\), \(d = k\) and \(n = 5\) or \(a = k\), \(l = 5k\) and \(n = 5\) AND sets equal to 165

M1: Attempts to solve their linear equation in \(k\) having set the sum of their first 5 terms equal to 165. Solving \(V_5 = 165\) scores no marks.

A1: cao and cso