| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with logarithmic terms |
| Difficulty | Challenging +1.2 This is a structured method of differences question with clear scaffolding. Part (a) requires recognizing that ln((r+1)/(r-1)) = ln(r+1) - ln(r-1) and telescoping the sum, which is a standard Core Pure 1 technique. Part (b) applies the result with straightforward arithmetic manipulation using log laws. While it requires careful bookkeeping and understanding of telescoping series, it follows a well-established method without requiring novel insight, making it moderately above average difficulty for A-level. |
| Spec | 4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Applies \(\ln\!\left(\dfrac{r+1}{r-1}\right) = \ln(r+1) - \ln(r-1)\) to apply differences | M1 | 3.1a — uses subtraction law of logs |
| Expands telescoping sum showing cancellation, e.g. \((\ln 3 - \ln 1)+(\ln 4 - \ln 2)+\ldots+(\ln(n+1)-\ln(n-1))\) | dM1 | 1.1b — minimum of \(r=2,3,4,\ldots,n-1,n\) shown; at least one cancelling term established |
| \(\ln n + \ln(n+1) - \ln 2\) | A1 | 1.1b — correct non-cancelling terms |
| \(\ln\!\left(\dfrac{n(n+1)}{2}\right)\) | A1* | 2.1 — printed answer; no errors or omissions; complete list required before extraction |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\displaystyle\sum_{r=51}^{100}\ln\!\left(\dfrac{r+1}{r-1}\right) = \sum_{r=2}^{100}\ln\!\left(\dfrac{r+1}{r-1}\right) - \sum_{r=2}^{50}\ln\!\left(\dfrac{r+1}{r-1}\right) = \ln\!\left(\dfrac{100\times101}{2}\right) - \ln\!\left(\dfrac{50\times51}{2}\right)\) | M1 | 1.1b — splits into sum to 100 minus sum to \(k\) (\(k=49,50,51\)); applies (a) |
| Uses power log law: \(35\ln\!\left(\dfrac{100\times101}{2} \div \dfrac{50\times51}{2}\right)\) | M1 | 3.1a — applies difference and power log laws correctly |
| \(35\ln\!\left(\dfrac{202}{51}\right)\) | A1 | 1.1b — accept \(35\ln\dfrac{5050}{1275}\) |
# Question 4:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| Applies $\ln\!\left(\dfrac{r+1}{r-1}\right) = \ln(r+1) - \ln(r-1)$ to apply differences | M1 | 3.1a — uses subtraction law of logs |
| Expands telescoping sum showing cancellation, e.g. $(\ln 3 - \ln 1)+(\ln 4 - \ln 2)+\ldots+(\ln(n+1)-\ln(n-1))$ | dM1 | 1.1b — minimum of $r=2,3,4,\ldots,n-1,n$ shown; at least one cancelling term established |
| $\ln n + \ln(n+1) - \ln 2$ | A1 | 1.1b — correct non-cancelling terms |
| $\ln\!\left(\dfrac{n(n+1)}{2}\right)$ | A1* | 2.1 — printed answer; no errors or omissions; complete list required before extraction |
## Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\displaystyle\sum_{r=51}^{100}\ln\!\left(\dfrac{r+1}{r-1}\right) = \sum_{r=2}^{100}\ln\!\left(\dfrac{r+1}{r-1}\right) - \sum_{r=2}^{50}\ln\!\left(\dfrac{r+1}{r-1}\right) = \ln\!\left(\dfrac{100\times101}{2}\right) - \ln\!\left(\dfrac{50\times51}{2}\right)$ | M1 | 1.1b — splits into sum to 100 minus sum to $k$ ($k=49,50,51$); applies (a) |
| Uses power log law: $35\ln\!\left(\dfrac{100\times101}{2} \div \dfrac{50\times51}{2}\right)$ | M1 | 3.1a — applies difference and power log laws correctly |
| $35\ln\!\left(\dfrac{202}{51}\right)$ | A1 | 1.1b — accept $35\ln\dfrac{5050}{1275}$ |\begin{enumerate}
\item (a) Use the method of differences to prove that for $n > 2$
\end{enumerate}
$$\sum _ { r = 2 } ^ { n } \ln \left( \frac { r + 1 } { r - 1 } \right) \equiv \ln \left( \frac { n ( n + 1 ) } { 2 } \right)$$
(4)\\
(b) Hence find the exact value of
$$\sum _ { r = 51 } ^ { 100 } \ln \left( \frac { r + 1 } { r - 1 } \right) ^ { 35 }$$
Give your answer in the form $a \ln \left( \frac { b } { c } \right)$ where $a , b$ and $c$ are integers to\\
be determined.
\hfill \mbox{\textit{Edexcel CP1 2022 Q4 [7]}}