| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Standard summation formulae application |
| Difficulty | Moderate -0.3 Part (a) requires routine expansion and application of standard summation formulae (∑r and ∑r²), which is straightforward algebra. Part (b) is a standard induction proof with no tricky algebraic manipulation in the inductive step. While this is Further Maths content, both parts are textbook exercises requiring only methodical application of techniques with no problem-solving insight needed. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | 1 1 1 |
| Answer | Marks |
|---|---|
| as true for n = 1, true for all n | B1 |
| Answer | Marks |
|---|---|
| [6] | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | checking n = 1 |
Question 6:
6 | (b) | 1 1 1
r ( r + 2 ) = 1 3 = 3 , n ( n + 1 ) ( 2 n + 7 ) = 1 2 9 = 3
6 6
=r 1
so true for n = 1
Assume true for n = k
k+1 1
r(r+2)= k(k+1)(2k+7)+(k+1)(k+3)
6
r=1
1 1
= ( k + 1 )( 2 k 2 + 7 k + 6 k + 1 8 ) = ( k + 1 )( 2 k 2 + 1 3 k + 1 8 )
6 6
1
= ( k + 1 )( k + 2 )( 2 k + 9 )
6
1
= (k+1)(k+1+1)(2(k+1)+7)
6
so if true for n = k then true for n = k + 1
as true for n = 1, true for all n | B1
M1
M1
A1*
A1
A1cao
[6] | 1.1
2.1
1.1
2.1
2.2a
2.4 | checking n = 1
+k 1
condone k ( k + 2 ) = ...
=r 1
factorising
or equals target expression if given
dep A1* must have both statements6
\begin{enumerate}[label=(\alph*)]
\item Using standard summation formulae, show that $\sum _ { r = 1 } ^ { n } r ( r + 2 ) = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 7 )$.
\item Use induction to prove the result in part (a).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2022 Q6 [10]}}