A2 June 2023 Paper 1 Q5

AQACurrent spec6 marksSeries

5 The function \(\mathrm{f}\) is defined by

\[\mathrm{f}(r) = 2^r(r - 2) \qquad (r \in \mathbb{Z})\]

(a) Show that\[\mathrm{f}(r + 1) - \mathrm{f}(r) = r2^r\] [2 marks]

(b) Use the method of differences to show that\[\sum_{r=1}^{n} r2^r = 2^{n+1}(n - 1) + 2\] [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Obtains a correct expression for \(\mathrm{f}(r + 1)\) and forms an expression for the difference \(\mathrm{f}(r + 1) - \mathrm{f}(r)\) Eg \(2^{r+1}(r - 1) - 2^r(r - 2)\)	M1	1.1a
Completes a rigorous argument to obtain the required result Must include \(\mathrm{f}(r + 1) - \mathrm{f}(r)\), \(r2^r\) and at least one intermediate step between \(2^{r+1}(r - 1) - 2^r(r - 2)\) and \(r2^r\)	R1	2.1
	(2)

Typical solution

\[\begin{aligned} \mathrm{f}(r + 1) - \mathrm{f}(r) &= 2^{r+1}(r - 1) - 2^r(r - 2) \\ &= 2^r\big(2(r - 1) - (r - 2)\big) \\ &= 2^r(2r - 2 - r + 2) \\ &= r2^r \end{aligned}\]

Mark scheme (b)

Scheme	Marks	AO
Writes at least two lines of subtracting terms, using part (a) Allow \(\mathrm{f}(2) - \mathrm{f}(1)\) etc	M1	1.1a
Writes at least two consecutive lines showing cancellation (PI)	M1	2.5
Correctly reduces the expression to two terms	A1	1.1b
Completes a reasoned argument using the method of differences to reach the required result. This mark is only available if at least the first two lines and the last two lines, to reach the required result, are seen.	R1	2.1
	(4)
	(6 marks)

Typical solution

\[\begin{aligned} \sum_{r=1}^{n} r2^r &= \sum_{r=1}^{n} \mathrm{f}(r + 1) - \mathrm{f}(r) \\ &= \cancel{0} - (-2) \\ &\quad + \cancel{8} - \cancel{0} \\ &\quad + \cancel{32} - \cancel{8} \\ &\quad + \ldots \\ &\quad + \cancel{2^n(n - 2)} - \cancel{2^{n-1}(n - 3)} \\ &\quad + 2^{n+1}(n - 1) - \cancel{2^n(n - 2)} \end{aligned}\]\[\begin{aligned} \sum_{r=1}^{n} r2^r &= 2^{n+1}(n - 1) - 2(-1) \\ &= 2^{n+1}(n - 1) + 2 \end{aligned}\]