| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure Core (Further Pure Core) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Proving standard summation formulae |
| Difficulty | Standard +0.3 This is a standard Further Maths technique using telescoping sums to derive summation formulae. Part (a) follows a well-established method with clear guidance ('by considering...'), and part (b) is a straightforward algebraic manipulation of the result. While it requires more sophistication than basic A-level, it's a routine exercise for Further Maths students with no novel problem-solving required. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | (𝑟+1)3−𝑟3 = 𝑟3+3𝑟2+3𝑟+1−𝑟3 |
| Answer | Marks |
|---|---|
| = (𝑛+1)3−1 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | Soi by M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (b) | 𝑛 |
| Answer | Marks |
|---|---|
| 3 | M1* |
| Answer | Marks |
|---|---|
| [4] | 2.5 |
| Answer | Marks |
|---|---|
| 2.2a | ∑𝑛 1 = 𝑛used and starting to rearrange |
Question 1:
1 | (a) | (𝑟+1)3−𝑟3 = 𝑟3+3𝑟2+3𝑟+1−𝑟3
= 3𝑟2+3𝑟+1
𝑛 𝑛
∑(3𝑟2+3𝑟+1) = ∑[(𝑟+1)3−𝑟3]
𝑟=1 𝑟=1
=23−13+33−23+⋯+𝑛3−(𝑛−1)3+(𝑛+1)3
−𝑛3
= (𝑛+1)3−1 | B1
M1
A1
[3] | 1.1
2.5
2.2a | Soi by M1A1
Cannot use standard summation formulae for final 2 marks
isw
1 | (b) | 𝑛
3∑𝑟(𝑟+1)+𝑛 = (𝑛+1)3−1
𝑟=1
𝑛
1
⇒ ∑𝑟(𝑟+1) = [(𝑛+1)3−1−𝑛]
3
𝑟=1
1
= (𝑛+1)[(𝑛+1)2−1]
3
1
= 𝑛(𝑛+1)(𝑛+2)
3 | M1*
A1
M1dep
*
A1
[4] | 2.5
3.1a
1.1
2.2a | ∑𝑛 1 = 𝑛used and starting to rearrange
𝑟=1
Factorising with n or n + 1 correctly
Allow SC2 for correct solution using standard summation
formulae.1
\begin{enumerate}[label=(\alph*)]
\item By considering $( r + 1 ) ^ { 3 } - r ^ { 3 }$, find $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \left( 3 \mathrm { r } ^ { 2 } + 3 \mathrm { r } + 1 \right)$.
\item Use this result to find $\sum _ { r = 1 } ^ { n } r ( r + 1 )$, expressing your answer in fully factorised form.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure Core 2022 Q1 [7]}}