| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Method of differences with given identity |
| Difficulty | Standard +0.3 This is a standard Further Maths method of differences question with straightforward partial fractions (three linear factors with constant numerators). The telescoping is routine once the partial fractions are found, and the algebraic simplification to match the given form is mechanical. Slightly easier than average due to the structured guidance and standard technique application. |
| Spec | 4.05c Partial fractions: extended to quadratic denominators4.06b Method of differences: telescoping series |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{(2n-1)(2n+1)(2n+3)} \equiv \frac{A}{2n-1} + \frac{B}{2n+1} + \frac{C}{2n+3}\) | M1 | Correct partial fraction attempt to obtain values for \(A\), \(B\), \(C\) |
| \(\frac{1}{8(2n-1)} - \frac{1}{4(2n+1)} + \frac{1}{8(2n+3)}\) | A1 | Correct partial fractions (mark is for correct fractions, not just values of \(A,B,C\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses method of differences showing sufficient rows to establish cancelling (e.g. 3 rows at start and 2 at end) | M1 | May be implied by extracting correct non-cancelling terms |
| \(= \frac{1}{8}\left(1 - \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+3}\right)\) | A1 | Identifies correct non-cancelling terms; may be unsimplified |
| Attempts to combine into one fraction (at least one constant term and at least 2 different algebraic terms, at least 3 terms in numerator) | dM1 | Dependent on previous M1 |
| \(= \frac{n(n+2)}{3(2n+1)(2n+3)}\) | A1 | cao |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{(2n-1)(2n+1)(2n+3)} \equiv \frac{A}{2n-1} + \frac{B}{2n+1} + \frac{C}{2n+3}$ | M1 | Correct partial fraction attempt to obtain values for $A$, $B$, $C$ |
| $\frac{1}{8(2n-1)} - \frac{1}{4(2n+1)} + \frac{1}{8(2n+3)}$ | A1 | Correct partial fractions (mark is for correct fractions, not just values of $A,B,C$) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses method of differences showing sufficient rows to establish cancelling (e.g. 3 rows at start and 2 at end) | M1 | May be implied by extracting correct non-cancelling terms |
| $= \frac{1}{8}\left(1 - \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+3}\right)$ | A1 | Identifies correct non-cancelling terms; may be unsimplified |
| Attempts to combine into one fraction (at least one constant term and at least 2 different algebraic terms, at least 3 terms in numerator) | dM1 | Dependent on previous M1 |
| $= \frac{n(n+2)}{3(2n+1)(2n+3)}$ | A1 | cao |
---\begin{enumerate}
\item (a) Express
\end{enumerate}
$$\frac { 1 } { ( 2 n - 1 ) ( 2 n + 1 ) ( 2 n + 3 ) }$$
in partial fractions.\\
(b) Hence, using the method of differences, show that for all integer values of $n$,
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { n ( n + 2 ) } { a ( 2 n + b ) ( 2 n + c ) }$$
where $a$, $b$ and $c$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F2 2023 Q2 [6]}}