June 2018 Paper 2 Q4

EdexcelCurrent spec7 marksSequences & Series

(i) Show that \(\displaystyle\sum_{r=1}^{16}\left(3 + 5r + 2^r\right) = 131\,798\) (4)

(ii) A sequence \(u_1, u_2, u_3, \ldots\) is defined by\[u_{n+1} = \frac{1}{u_n}, \qquad u_1 = \frac{2}{3}\]

Find the exact value of \(\displaystyle\sum_{r=1}^{100}u_r\) (3)

Mark scheme (i)

Scheme	Marks	AO
Way 1: \(\left\{\displaystyle\sum_{r=1}^{16}\left(3 + 5r + 2^r\right) =\right\}\ \displaystyle\sum_{r=1}^{16}(3 + 5r) + \displaystyle\sum_{r=1}^{16}\left(2^r\right)\)	M1	3.1a
\(= \dfrac{16}{2}(2(8) + 15(5)) + \dfrac{2\left(2^{16} - 1\right)}{2 - 1}\)	M1 M1	1.1b 1.1b
\(= 728 + 131\,070 = 131\,798\) *	A1*	2.1
	(4)

Notes

(i) Way 2

Scheme	Marks	AO
\(\left\{\displaystyle\sum_{r=1}^{16}\left(3 + 5r + 2^r\right) =\right\}\ \displaystyle\sum_{r=1}^{16}3 + \displaystyle\sum_{r=1}^{16}(5r) + \displaystyle\sum_{r=1}^{16}\left(2^r\right)\)	M1	3.1a
\(= (3 \times 16) + \dfrac{16}{2}(2(5) + 15(5)) + \dfrac{2\left(2^{16} - 1\right)}{2 - 1}\)	M1 M1	1.1b 1.1b
\(= 48 + 680 + 131\,070 = 131\,798\) *	A1*	2.1
	(4)

(i) Way 3

Scheme	Marks	AO
Sum \(= 10 + 17 + 26 + 39 + 60 + 97 + 166 + 299 + 560 + 1077 + 2106\) \(+ 4159 + 8260 + 16457 + 32846 + 65619 = 131\,798\) *	M1 M1 M1 A1*	3.1a 1.1b 1.1b 2.1
	(4)

M1: Uses a correct methodical strategy to enable the given sum, \(\displaystyle\sum_{r=1}^{16}\left(3 + 5r + 2^r\right)\) to be found
Allow M1 for any of the following:

expressing the given sum as either
\(\displaystyle\sum_{r=1}^{16}(3 + 5r) + \displaystyle\sum_{r=1}^{16}\left(2^r\right),\ \ \displaystyle\sum_{r=1}^{16}3 + \displaystyle\sum_{r=1}^{16}(5r) + \displaystyle\sum_{r=1}^{16}\left(2^r\right)\) or \(\displaystyle\sum_{r=1}^{16}3 + 5\displaystyle\sum_{r=1}^{16}r + \displaystyle\sum_{r=1}^{16}\left(2^r\right)\)
attempting to find both \(\displaystyle\sum_{r=1}^{16}(3 + 5r)\) and \(\displaystyle\sum_{r=1}^{16}\left(2^r\right)\) separately
\((3 \times 16)\) and attempting to find both \(\displaystyle\sum_{r=1}^{16}(5r)\) and \(\displaystyle\sum_{r=1}^{16}\left(2^r\right)\) separately

M1: Way 1: Correct method for finding the sum of an AP with \(a = 8, d = 5, n = 16\)
Way 2: \((3 \times 16)\) and a correct method for finding the sum of an AP

M1: Correct method for finding the sum of a GP with \(a = 2, r = 2, n = 16\)

A1*: For all steps fully shown (with correct formulae used) leading to 131 798

Note: Way 1: Give 2^nd M1 for writing \(\displaystyle\sum_{r=1}^{16}(3 + 5r)\) as \(\dfrac{16}{2}(8 + 83)\)

Note: Way 2: Give 2^nd M1 for writing \(\displaystyle\sum_{r=1}^{16}3 + \displaystyle\sum_{r=1}^{16}(5r)\) as \(48 + \dfrac{16}{2}(5 + 80)\) or \(48 + 680\)

Note: Give 3^rd M1 for writing \(\displaystyle\sum_{r=1}^{16}\left(2^r\right)\) as \(\dfrac{2\left(1 - 2^{16}\right)}{1 - 2}\) or \(2\left(2^{16} - 1\right)\) or \(\left(2^{17} - 2\right)\)

(i) Way 3

M1: At least 6 correct terms and 16 terms shown

M1: At least 10 correct terms (may not be 16 terms)

M1: At least 15 correct terms (may not be 16 terms)

A1*: All 16 terms correct and an indication that the sum is 131 798

Mark scheme (ii)

Scheme	Marks	AO
\(\left\{u_1 = \dfrac{2}{3}\right\},\ u_2 = \dfrac{3}{2},\ u_3 = \dfrac{2}{3}, \ldots\) *(can be implied by later working)*	M1	1.1b
\(\left\{\displaystyle\sum_{r=1}^{100}u_r =\right\}\ 50\left(\dfrac{2}{3}\right) + 50\left(\dfrac{3}{2}\right)\) or \(50\left(\dfrac{2}{3} + \dfrac{3}{2}\right)\)	M1	2.2a
\(= \dfrac{325}{3}\ \left(\text{or } 108\dfrac{1}{3} \text{ or } 108.\dot{3} \text{ or } \dfrac{1300}{12} \text{ or } \dfrac{650}{6}\right)\)	A1	1.1b
	(3)
	(7 marks)

Notes

M1: For some indication that the next two terms of this sequence are \(\dfrac{3}{2}, \dfrac{2}{3}\)

M1: For deducing that the sum can be found by applying \(50\left(\dfrac{2}{3}\right) + 50\left(\dfrac{3}{2}\right)\) or \(50\left(\dfrac{2}{3} + \dfrac{3}{2}\right)\), o.e.

A1: Obtains \(\dfrac{325}{3}\) or \(108\dfrac{1}{3}\) or \(108.\dot{3}\) or an exact equivalent

Note: Allow 1^st M1 for \(u_2 = \dfrac{3}{2}\) (or equivalent) and \(u_3 = \dfrac{2}{3}\) (or equivalent)

Note: Allow 1^st M1 for the first 3 terms written as \(\dfrac{2}{3}, \dfrac{3}{2}, \dfrac{2}{3}, \ldots\)

Note: Allow 1^st M1 for the 2^nd and 3^rd terms written as \(\dfrac{3}{2}, \dfrac{2}{3}, \ldots\) in the correct order

Note: Condone \(\dfrac{2}{3}\) written as 0.66 or awrt 0.67 for the 1^st M1 mark

Note: Give A0 for 108.3 or 108.333... without reference to \(\dfrac{325}{3}\) or \(108\dfrac{1}{3}\) or \(108.\dot{3}\)