| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Standard summation formula application |
| Difficulty | Standard +0.3 This is a straightforward application of standard summation formulas with algebraic manipulation. Part (a) requires expanding (r+2)(r+1)r, applying three given formulas, and factorizing—routine for AS level. Part (b) involves equating two expressions and solving a quartic that likely factors nicely. Slightly above average due to the algebraic manipulation required, but no novel insight needed. |
| Spec | 1.04g Sigma notation: for sums of series4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume \(= r\times(r+1)\times(r+2)\) | B1 | 1.1b |
| \(\displaystyle\sum_{r=1}^{n}(r^3+3r^2+2r)\) | M1 | 3.1b; Complete method for total volume of \(n\) blocks in sigma notation |
| \(V=\frac{1}{4}n^2(n+1)^2+3\times\frac{1}{6}n(n+1)(2n+1)+2\times\frac{1}{2}n(n+1)\) | M1 | 2.1; Substitutes at least one standard formula |
| \(V=\frac{1}{4}n(n+1)\bigl[n(n+1)+2(2n+1)+4\bigr]\) | dM1 | 1.1b; Attempts to factorise \(\frac{1}{4}n(n+1)\); each term must contain factor \(n(n+1)\) |
| \(V=\frac{1}{4}n(n+1)(n^2+5n+6)\) \(\Rightarrow V=\frac{1}{4}n(n+1)(n+2)(n+3)\) * | A1* | 1.1b; No errors, no bracketing errors; going from \(\frac{1}{4}n(n^3+6n^2+11n+6)\) directly scores dM0 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\frac{1}{4}n(n+1)(n+2)(n+3)=n^4+6n^3-11710\); simplifies to \(3n^4+18n^3-11n^2-6n-46840=0\) and solves | M1 | 1.1b |
| There are 10 blocks, \(n=10\) | A1 | 3.2a; Must select \(n=10\) from a correct equation |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume $= r\times(r+1)\times(r+2)$ | B1 | 1.1b |
| $\displaystyle\sum_{r=1}^{n}(r^3+3r^2+2r)$ | M1 | 3.1b; Complete method for total volume of $n$ blocks in sigma notation |
| $V=\frac{1}{4}n^2(n+1)^2+3\times\frac{1}{6}n(n+1)(2n+1)+2\times\frac{1}{2}n(n+1)$ | M1 | 2.1; Substitutes at least one standard formula |
| $V=\frac{1}{4}n(n+1)\bigl[n(n+1)+2(2n+1)+4\bigr]$ | dM1 | 1.1b; Attempts to factorise $\frac{1}{4}n(n+1)$; each term must contain factor $n(n+1)$ |
| $V=\frac{1}{4}n(n+1)(n^2+5n+6)$ $\Rightarrow V=\frac{1}{4}n(n+1)(n+2)(n+3)$ * | A1* | 1.1b; No errors, no bracketing errors; going from $\frac{1}{4}n(n^3+6n^2+11n+6)$ directly scores dM0 A0 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\frac{1}{4}n(n+1)(n+2)(n+3)=n^4+6n^3-11710$; simplifies to $3n^4+18n^3-11n^2-6n-46840=0$ and solves | M1 | 1.1b |
| There are 10 blocks, $n=10$ | A1 | 3.2a; Must select $n=10$ from a correct equation |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{09bd14c0-c368-4ae1-bee0-cc8bf82abecc-12_351_655_246_705}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A block has length $( r + 2 ) \mathrm { cm }$, width $( r + 1 ) \mathrm { cm }$ and height $r \mathrm {~cm}$, as shown in Figure 2.\\
In a set of $n$ such blocks, the first block has a height of 1 cm , the second block has a height of 2 cm , the third block has a height of 3 cm and so on.
\begin{enumerate}[label=(\alph*)]
\item Use the standard results for $\sum _ { r = 1 } ^ { n } r ^ { 3 } , \sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r$ to show that the total volume, $V$, of all $n$ blocks in the set is given by
$$V = \frac { n } { 4 } ( n + 1 ) ( n + 2 ) ( n + 3 ) \quad n \geqslant 1$$
Given that the total volume of all $n$ blocks is
$$\left( n ^ { 4 } + 6 n ^ { 3 } - 11710 \right) \mathrm { cm } ^ { 3 }$$
\item determine how many blocks make up the set.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP AS 2020 Q5 [7]}}