| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Simple recurrence evaluation |
| Difficulty | Moderate -0.3 Part (i) is straightforward summation using standard formulas for arithmetic and geometric series. Part (ii) requires recognizing the recurrence creates a period-2 cycle (u₁=2/3, u₂=3/2, u₃=2/3,...), then summing 50 pairs. This is easier than average as the pattern recognition is immediate and the arithmetic is simple, though it does require some problem-solving insight rather than pure recall. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series1.04h Arithmetic sequences: nth term and sum formulae1.04i Geometric sequences: nth term and finite series sum |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{16}(3+5r+2^r) = \sum_{r=1}^{16}(3+5r) + \sum_{r=1}^{16}(2^r)\) | M1 | 3.1a |
| \(= \frac{16}{2}(2(8)+15(5))\) | M1 | 1.1b - Correct method for AP with \(a=8\), \(d=5\), \(n=16\) |
| \(+\frac{2(2^{16}-1)}{2-1}\) | M1 | 1.1b - Correct method for GP with \(a=2\), \(r=2\), \(n=16\) |
| \(= 728 + 131070 = 131798\) | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sum_{r=1}^{16}(3+5r+2^r) = \sum_{r=1}^{16}3 + \sum_{r=1}^{16}(5r) + \sum_{r=1}^{16}(2^r)\) | M1 | 3.1a |
| \(= (3\times16) + \frac{16}{2}(2(5)+15(5))\) | M1 | 1.1b |
| \(+\frac{2(2^{16}-1)}{2-1}\) | M1 | 1.1b |
| \(= 48 + 680 + 131070 = 131798\) | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(u_1=\frac{2}{3}\), \(u_2=\frac{3}{2}\), \(u_3=\frac{2}{3}\), ... (can be implied by later working) | M1 | 1.1b |
| \(\left\{\sum_{r=1}^{100}u_r =\right\} 50\!\left(\frac{2}{3}\right)+50\!\left(\frac{3}{2}\right)\) or \(50\!\left(\frac{2}{3}+\frac{3}{2}\right)\) | M1 | 2.2a |
| \(= \frac{325}{3}\) (or \(108\frac{1}{3}\) or \(108.\dot{3}\) or \(\frac{1300}{12}\) or \(\frac{650}{6}\)) | A1 | 1.1b |
# Question 4:
## Part (i) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{16}(3+5r+2^r) = \sum_{r=1}^{16}(3+5r) + \sum_{r=1}^{16}(2^r)$ | M1 | 3.1a |
| $= \frac{16}{2}(2(8)+15(5))$ | M1 | 1.1b - Correct method for AP with $a=8$, $d=5$, $n=16$ |
| $+\frac{2(2^{16}-1)}{2-1}$ | M1 | 1.1b - Correct method for GP with $a=2$, $r=2$, $n=16$ |
| $= 728 + 131070 = 131798$ | A1* | 2.1 |
## Part (i) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sum_{r=1}^{16}(3+5r+2^r) = \sum_{r=1}^{16}3 + \sum_{r=1}^{16}(5r) + \sum_{r=1}^{16}(2^r)$ | M1 | 3.1a |
| $= (3\times16) + \frac{16}{2}(2(5)+15(5))$ | M1 | 1.1b |
| $+\frac{2(2^{16}-1)}{2-1}$ | M1 | 1.1b |
| $= 48 + 680 + 131070 = 131798$ | A1* | 2.1 |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u_1=\frac{2}{3}$, $u_2=\frac{3}{2}$, $u_3=\frac{2}{3}$, ... (can be implied by later working) | M1 | 1.1b |
| $\left\{\sum_{r=1}^{100}u_r =\right\} 50\!\left(\frac{2}{3}\right)+50\!\left(\frac{3}{2}\right)$ or $50\!\left(\frac{2}{3}+\frac{3}{2}\right)$ | M1 | 2.2a |
| $= \frac{325}{3}$ (or $108\frac{1}{3}$ or $108.\dot{3}$ or $\frac{1300}{12}$ or $\frac{650}{6}$) | A1 | 1.1b |
---\begin{enumerate}
\item (i) Show that $\sum _ { r = 1 } ^ { 16 } \left( 3 + 5 r + 2 ^ { r } \right) = 131798$\\
(ii) A sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is defined by
\end{enumerate}
$$u _ { n + 1 } = \frac { 1 } { u _ { n } } , \quad u _ { 1 } = \frac { 2 } { 3 }$$
Find the exact value of $\sum _ { r = 1 } ^ { 100 } u _ { r }$
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q4 [7]}}