June 2019 Paper 2 Q8

EdexcelCurrent spec6 marksLogs & Exponentials Sequences & Series

(i) Find the value of\[\displaystyle\sum_{r=4}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r\] (3)

(ii) Show that\[\displaystyle\sum_{n=1}^{48} \log_5\left(\dfrac{n+2}{n+1}\right) = 2\] (3)

Mark scheme (i)

Scheme	Marks	AO
Way 1 \(\displaystyle\sum_{r=4}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r = 20\left(\dfrac{1}{2}\right)^4 + 20\left(\dfrac{1}{2}\right)^5 + 20\left(\dfrac{1}{2}\right)^6 + \ldots\)
\(= \dfrac{20\left(\frac{1}{2}\right)^4}{1 - \frac{1}{2}}\)	M1 M1	1.1b 3.1a
\(\{= (1.25)(2)\} = 2.5\) o.e.	A1	1.1b
	(3)

Notes

(i) Way 2

Scheme	Marks	AO
\(\displaystyle\sum_{r=4}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r = \displaystyle\sum_{r=1}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r - \displaystyle\sum_{r=1}^{3} 20 \times \left(\dfrac{1}{2}\right)^r\)
\(= \dfrac{10}{1 - \frac{1}{2}} - (10 + 5 + 2.5)\) or \(= \dfrac{10}{1 - \frac{1}{2}} - \dfrac{10\left(1 - \left(\tfrac{1}{2}\right)^3\right)}{1 - \frac{1}{2}}\)	M1 M1	1.1b 3.1a
\(\{= 20 - 17.5\} = 2.5\) o.e.	A1	1.1b
	(3)

(i) Way 3

Scheme	Marks	AO
\(\displaystyle\sum_{r=4}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r = \displaystyle\sum_{r=0}^{\infty} 20 \times \left(\dfrac{1}{2}\right)^r - \displaystyle\sum_{r=0}^{3} 20 \times \left(\dfrac{1}{2}\right)^r\)
\(= \dfrac{20}{1 - \frac{1}{2}} - (20 + 10 + 5 + 2.5)\) or \(= \dfrac{20}{1 - \frac{1}{2}} - \dfrac{20\left(1 - \left(\tfrac{1}{2}\right)^4\right)}{1 - \frac{1}{2}}\)	M1 M1	1.1b 3.1a
\(\{= 40 - 37.5\} = 2.5\) o.e.	A1	1.1b
	(3)

Way 1

M1: Applies \(\dfrac{a}{1-r}\) for their \(r\) (where \(-1 \lt\) their \(r \lt 1\)) and their value for \(a\)

M1: Finds the infinite sum by using a complete strategy of applying \(\dfrac{20\left(\frac{1}{2}\right)^4}{1 - \frac{1}{2}}\)

A1: 2.5 o.e.

Way 2

M1: Applies \(\dfrac{a}{1-r}\) for their \(r\) (where \(-1 \lt\) their \(r \lt 1\)) and their value for \(a\)

M1: Finds the infinite sum by using a completely correct strategy of applying\[\dfrac{10}{1 - \frac{1}{2}} - (10 + 5 + 2.5) \quad \text{or} \quad \dfrac{10}{1 - \frac{1}{2}} - \dfrac{10\left(1 - \left(\tfrac{1}{2}\right)^3\right)}{1 - \frac{1}{2}}\]

A1: 2.5 o.e.

Way 3

M1: Applies \(\dfrac{a}{1-r}\) for their \(r\) (where \(-1 \lt\) their \(r \lt 1\)) and their value for \(a\)

M1: Finds the infinite sum by using a completely correct strategy of applying\[\dfrac{20}{1 - \frac{1}{2}} - (20 + 10 + 5 + 2.5) \quad \text{or} \quad \dfrac{20}{1 - \frac{1}{2}} - \dfrac{20\left(1 - \left(\tfrac{1}{2}\right)^4\right)}{1 - \frac{1}{2}}\]

A1: 2.5 o.e.

Note: Give M1 M1 A1 for a correct answer of 2.5 from no working in (i)

Mark scheme (ii)

Scheme	Marks	AO
Way 1 \(\left\{\displaystyle\sum_{n=1}^{48} \log_5\left(\dfrac{n+2}{n+1}\right) =\right\}\)
\(= \log_5\left(\dfrac{3}{2}\right) + \log_5\left(\dfrac{4}{3}\right) + \ldots\ldots + \log_5\left(\dfrac{50}{49}\right) = \log_5\left(\dfrac{3}{2} \times \dfrac{4}{3} \times \ldots \times \dfrac{50}{49}\right)\)	M1 M1	1.1b 3.1a
\(= \log_5\left(\dfrac{50}{2}\right)\) or \(\log_5(25) = 2\) *	A1*	2.1
	(3)
	(6 marks)

Notes

(ii) Way 2

Scheme	Marks	AO
\(\left\{\displaystyle\sum_{n=1}^{48} \log_5\left(\dfrac{n+2}{n+1}\right) =\right\}\ \displaystyle\sum_{n=1}^{48} \left(\log_5(n+2) - \log_5(n+1)\right)\)	M1	1.1b
\(= (\log_5 3 + \log_5 4 + \ldots\ldots + \log_5 50) - (\log_5 2 + \log_5 3 + \ldots\ldots + \log_5 49)\)	M1	3.1a
\(= \log_5 50 - \log_5 2\) or \(\log_5\left(\dfrac{50}{2}\right)\) or \(\log_5(25) = 2\) *	A1*	2.1
	(3)

Way 1

M1: Some evidence of applying the addition law of logarithms as part of a valid proof

M1: Begins to solve the problem by just writing (or by combining) at least three terms including

either the first two terms and the last term
or the first term and the last two terms

Note: The 2nd mark can be gained by writing any of

listing \(\log_5\left(\dfrac{3}{2}\right), \log_5\left(\dfrac{4}{3}\right), \log_5\left(\dfrac{50}{49}\right)\) or \(\log_5\left(\dfrac{3}{2}\right), \log_5\left(\dfrac{49}{48}\right), \log_5\left(\dfrac{50}{49}\right)\)
\(\log_5\left(\dfrac{3}{2}\right) + \log_5\left(\dfrac{4}{3}\right) + \ldots\ldots + \log_5\left(\dfrac{50}{49}\right)\)
\(\log_5\left(\dfrac{3}{2}\right) + \ldots\ldots + \log_5\left(\dfrac{49}{48}\right) + \log_5\left(\dfrac{50}{49}\right)\)
\(\log_5\left(\dfrac{3}{2} \times \dfrac{4}{3} \times \ldots \times \dfrac{50}{49}\right)\) {this will also gain the 1^st M1 mark}
\(\log_5\left(\dfrac{3}{2} \times \ldots \times \dfrac{49}{48} \times \dfrac{50}{49}\right)\) {this will also gain the 1^st M1 mark}

A1*: Correct proof leading to a correct answer of 2

Note: Do not allow the 2^nd M1 if \(\log_5\left(\dfrac{3}{2}\right), \log_5\left(\dfrac{4}{3}\right)\) are listed and \(\log_5\left(\dfrac{50}{49}\right)\) is used for the first time in their applying the formula \(S_{48} = \dfrac{48}{2}\left(\log_5\left(\dfrac{3}{2}\right) + \log_5\left(\dfrac{50}{49}\right)\right)\)

Note: Listing all 48 terms
Give M0 M1 A0 for \(\log_5\left(\dfrac{3}{2}\right) + \log_5\left(\dfrac{4}{3}\right) + \log_5\left(\dfrac{5}{4}\right) + \ldots\ldots + \log_5\left(\dfrac{50}{49}\right) = 2\) {lists all terms}
Give M0 M0 A0 for \(0.2519\ldots + 0.1787\ldots + 0.1386\ldots + \ldots\ldots + 0.0125\ldots = 2\) {all terms in decimals}

Way 2

M1: Uses the subtraction law of logarithms to give \(\log_5\left(\dfrac{n+2}{n+1}\right) \to \log_5(n+2) - \log_5(n+1)\)

M1: Begins to solve the problem by writing at least three terms for each of \(\log_5(n+2)\) and \(\log_5(n+1)\) including

either the first two terms and the last term for both \(\log_5(n+2)\) and \(\log_5(n+1)\)
or the first term and the last two terms for both \(\log_5(n+2)\) and \(\log_5(n+1)\)

Note: This mark can be gained by writing any of

\((\log_5 3 + \log_5 4 + \ldots\ldots + \log_5 50) - (\log_5 2 + \log_5 3 + \ldots\ldots + \log_5 49)\)
\((\log_5 3 + \ldots\ldots + \log_5 49 + \log_5 50) - (\log_5 2 + \ldots\ldots + \log_5 48 + \log_5 49)\)
\((\log_5 3 + \log_5 4 + \ldots\ldots + \log_5 50) - (\log_5 2 + \log_5 3 + \ldots\ldots + \log_5 49)\)
\((\log_5 3 - \log_5 2) + (\log_5 4 - \log_5 3) + \ldots\ldots + (\log_5 50 - \log_5 49)\)
\(\log_5 3 - \log_5 2,\ \ldots\ldots,\ \log_5 49 - \log_5 48,\ \log_5 50 - \log_5 49\)

A1*: Correct proof leading to a correct answer of 2

Both ways

Note: The base of 5 can be omitted for the M marks in part (ii), but the base of 5 must be included in the final line (as shown on the mark scheme) of their solution.

Note: If a student uses a mixture of a Way 1 or Way 2 method, then award the best Way 1 mark only or the best Way 2 mark only.

Note: Give M1 M0 A0 (1^st M for implied use of subtraction law of logarithms) for \(\displaystyle\sum_{n=1}^{48} \log_5\left(\dfrac{n+2}{n+1}\right) = 91.8237\ldots - 89.8237\ldots = 2\)

Note: Give M1 M1 A1 for\[\begin{aligned}\displaystyle\sum_{n=1}^{48} \log_5\left(\dfrac{n+2}{n+1}\right) &= \displaystyle\sum_{n=1}^{48} \left(\log_5(n+2) - \log_5(n+1)\right) \\ &= \log_5(3 \times 4 \times \ldots\ldots \times 50) - \log_5(2 \times 3 \times \ldots\ldots \times 49) \\ &= \log_5\left(\dfrac{50!}{2}\right) - \log_5(49!) \quad \text{or} \quad = \log_5(25 \times 49!) - \log_5(49!) \\ &= \log_5 25 = 2\end{aligned}\]