| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| 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 +0.3 This is a straightforward application of standard summation formulas requiring students to manipulate the sum from r=1 to 3n minus the sum from r=1 to 2n, then simplify algebraically. While it involves multiple steps and algebraic manipulation, it's a routine technique taught explicitly in FP1 with no novel insight required—slightly easier than average for Further Maths. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(\displaystyle\sum_{r=1}^{3n} r^2 = \dfrac{1}{6}3n(3n+1)(6n+1)\) or \(\dfrac{1}{2}n(3n+1)(6n+1)\) or equivalent | B1 | Either right hand side or exact equivalent - isw if expanded |
| (b) State/use \(\displaystyle\sum_{r=1}^{2n} r^2 = \dfrac{1}{3}n(2n+1)(4n+1)\) | B1 | States or uses this formula |
| Attempt \(\displaystyle\sum_{r=1}^{3n} r^2 - \sum_{r=1}^{2n} r^2 = \dfrac{n}{6}\{3(3n+1)(6n+1) - 2(2n+1)(4n+1)\}\) | M1 | Subtracts their sum to \(2n\) or \(2n-1\) and attempts to factorise by \(\dfrac{n}{6}\) seen anywhere |
| \(= \dfrac{n}{6}\{(54n^2 + 27n + 3) - (16n^2 + 12n + 2)\}\) | dM1 | Expands two quadratics; dependent on first M1 |
| \(= \dfrac{n}{6}\{38n^2 + 15n + 1\}\), so \(a=38, b=15, c=1\) | A1 | cao |
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $\displaystyle\sum_{r=1}^{3n} r^2 = \dfrac{1}{6}3n(3n+1)(6n+1)$ or $\dfrac{1}{2}n(3n+1)(6n+1)$ or equivalent | B1 | Either right hand side or exact equivalent - isw if expanded |
| **(b)** State/use $\displaystyle\sum_{r=1}^{2n} r^2 = \dfrac{1}{3}n(2n+1)(4n+1)$ | B1 | States or uses this formula |
| Attempt $\displaystyle\sum_{r=1}^{3n} r^2 - \sum_{r=1}^{2n} r^2 = \dfrac{n}{6}\{3(3n+1)(6n+1) - 2(2n+1)(4n+1)\}$ | M1 | Subtracts their sum to $2n$ or $2n-1$ **and** attempts to factorise by $\dfrac{n}{6}$ seen anywhere |
| $= \dfrac{n}{6}\{(54n^2 + 27n + 3) - (16n^2 + 12n + 2)\}$ | dM1 | Expands two quadratics; dependent on first M1 |
| $= \dfrac{n}{6}\{38n^2 + 15n + 1\}$, so $a=38, b=15, c=1$ | A1 | cao |
---\begin{enumerate}
\item (a) Using the formula for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ write down, in terms of $n$ only, an expression for
\end{enumerate}
$$\sum _ { r = 1 } ^ { 3 n } r ^ { 2 }$$
(b) Show that, for all integers $n$, where $n > 0$
$$\sum _ { r = 2 n + 1 } ^ { 3 n } r ^ { 2 } = \frac { n } { 6 } \left( a n ^ { 2 } + b n + c \right)$$
where the values of the constants $a$, $b$ and $c$ are to be found.\\
\hfill \mbox{\textit{Edexcel FP1 2016 Q3 [5]}}