| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Sum from n+1 to 2n or similar range |
| Difficulty | Standard +0.8 Part (a) is routine application of standard summation formulae. Part (b) requires recognizing the method of differences setup and partial fractions, which is standard Further Maths technique. Part (c) requires the insight to use results from (a) and (b) to find sum from n+1 to 2n by subtraction, involving algebraic manipulation. The multi-step nature and the non-standard summation range elevate this above average difficulty, but it follows established patterns for Further Maths questions. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6\left(\dfrac{1}{6}n(n+1)(2n+1)\right) + 6\left(\dfrac{1}{2}n(n+1)\right)[-5n]\) | M1 | Substitutes formulae for \(\sum r^2\) and \(\sum r\). |
| \(2n^3 + 6n^2 - n\) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{6r^2+6r-5}{r^2+r} = 6 - \frac{5}{r(r+1)}\) | B1 | Divides by denominator |
| \(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\) | B1 | Finds partial fractions |
| \(\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{1}{n} - \frac{1}{n+1}\) | M1 | Writes at least three complete terms, including the last term, to show cancellation |
| \(\sum_{r=1}^{n} \frac{6r^2+6r-5}{r^2+r} = 6n - 5 + \frac{5}{n+1}\) | A1 | OE e.g. \(\frac{6n^2+n}{n+1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sum_{r=1}^{2n} \frac{6r^2+6r-5}{r^2+r} - \sum_{r=1}^{n} \frac{6r^2+6r-5}{r^2+r} = 12n-5+\frac{5}{2n+1}-6n+5-\frac{5}{n+1}\) | M1 | Or uses method of differences again |
| \(12n-5+\frac{5}{2n+1}-6n+5-\frac{5}{n+1} = 6n + \frac{5}{2n+1} - \frac{5}{n+1}\) | A1 | Or \(6n - \frac{5n}{(n+1)(2n+1)}\). OE, like terms collected |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6\left(\dfrac{1}{6}n(n+1)(2n+1)\right) + 6\left(\dfrac{1}{2}n(n+1)\right)[-5n]$ | M1 | Substitutes formulae for $\sum r^2$ and $\sum r$. |
| $2n^3 + 6n^2 - n$ | A1 | |
| **Total** | **2** | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{6r^2+6r-5}{r^2+r} = 6 - \frac{5}{r(r+1)}$ | B1 | Divides by denominator |
| $\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$ | B1 | Finds partial fractions |
| $\sum_{r=1}^{n} \frac{1}{r(r+1)} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{1}{n} - \frac{1}{n+1}$ | M1 | Writes at least three complete terms, including the last term, to show cancellation |
| $\sum_{r=1}^{n} \frac{6r^2+6r-5}{r^2+r} = 6n - 5 + \frac{5}{n+1}$ | A1 | OE e.g. $\frac{6n^2+n}{n+1}$ |
## Question 2(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sum_{r=1}^{2n} \frac{6r^2+6r-5}{r^2+r} - \sum_{r=1}^{n} \frac{6r^2+6r-5}{r^2+r} = 12n-5+\frac{5}{2n+1}-6n+5-\frac{5}{n+1}$ | M1 | Or uses method of differences again |
| $12n-5+\frac{5}{2n+1}-6n+5-\frac{5}{n+1} = 6n + \frac{5}{2n+1} - \frac{5}{n+1}$ | A1 | Or $6n - \frac{5n}{(n+1)(2n+1)}$. OE, like terms collected |2
\begin{enumerate}[label=(\alph*)]
\item Use standard results from the list of formulae (MF19) to show that
$$\sum _ { r = 1 } ^ { n } \left( 6 r ^ { 2 } + 6 r - 5 \right) = a n ^ { 3 } + b n ^ { 2 } + c n$$
where $a$, $b$ and $c$ are integers to be determined.
\item Use the method of differences to find $\sum _ { r = 1 } ^ { n } \frac { 6 r ^ { 2 } + 6 r - 5 } { r ^ { 2 } + r }$ in terms of $n$.
\item Find also $\sum _ { r = n + 1 } ^ { 2 n } \frac { 6 r ^ { 2 } + 6 r - 5 } { r ^ { 2 } + r }$ in terms of $n$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q2 [8]}}