| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Arithmetic Sequences and Series |
| Type | Recurrence relation: evaluate sum |
| Difficulty | Standard +0.3 Part (i)(a) is trivial substitution. Part (i)(b) requires recognizing the recurrence produces a cycle (U₁=4, U₂=4, U₃=-4, U₄=-4, repeating), then summing 50 complete cycles. Part (ii) is a standard arithmetic series inequality requiring formula application and solving a quadratic inequality. While multi-step, all techniques are routine Core 1/2 material with no novel insight required. |
| Spec | 1.04e Sequences: nth term and recurrence relations1.04g Sigma notation: for sums of series |
| Answer | Marks | Guidance |
|---|---|---|
| \(U_2 = \frac{4}{4-3} = 4\) | B1 | States that \(U_2\) is 4. Accept \(\frac{4}{1}\) but not \(\frac{4}{4-3}\) unsimplified. Note \(U_1 = 4\) so do not award this B1. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{n=1}^{100} U_n = 100 \times 4 = 400\) | M1 A1 | M1: Uses method \(\sum U_n = k \times 4\) where \(k = 100\) or \(99\). Accept AP formula with \(a=4, d=0, n=99/100\). A1: 400. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n}(100-3r) < 0 \Rightarrow 97+94+91+\ldots+(100-3r) < 0\) | — | — |
| AP with \(a=97, d=-3, n=n\): \(S = \frac{n}{2}(2\times97+(n-1)\times-3) < 0\) | M1 | Uses \(S = \frac{n}{2}(2a+(n-1)d)\) with \(S=0\), \(a=97\) or \(100\), \(d=-3\). Not from linear equation (M0). |
| \(\Rightarrow \frac{n}{2}(197-3n) < 0 \Rightarrow n > 65.\dot{6}\) | dM1 | Dependent on previous M. Solving quadratic in \(n\). Accept \(n=, n>, n<\). |
| \(\Rightarrow n = 66\) | A1 | \(n=66\) cso |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n}(100-3r)<0 \Rightarrow \sum_{r=1}^{n}3r > \sum_{r=1}^{n}100\) | — | — |
| \(\Rightarrow 3\frac{n(n+1)}{2} > 100n\) | M1 M1A1 | M1: Attempting to solve using \(\sum r = \frac{n(n+1)}{2}\) and \(\sum 100 = 100n\). |
| \(\Rightarrow n > 65.6 \Rightarrow n = 66\) | — | — |
## Question 5(i)(a):
$U_2 = \frac{4}{4-3} = 4$ | B1 | States that $U_2$ is 4. Accept $\frac{4}{1}$ but not $\frac{4}{4-3}$ unsimplified. Note $U_1 = 4$ so do not award this B1.
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## Question 5(i)(b):
$\sum_{n=1}^{100} U_n = 100 \times 4 = 400$ | M1 A1 | M1: Uses method $\sum U_n = k \times 4$ where $k = 100$ or $99$. Accept AP formula with $a=4, d=0, n=99/100$. A1: 400.
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## Question 5(ii):
$\sum_{r=1}^{n}(100-3r) < 0 \Rightarrow 97+94+91+\ldots+(100-3r) < 0$ | — | —
AP with $a=97, d=-3, n=n$: $S = \frac{n}{2}(2\times97+(n-1)\times-3) < 0$ | M1 | Uses $S = \frac{n}{2}(2a+(n-1)d)$ with $S=0$, $a=97$ or $100$, $d=-3$. Not from linear equation (M0).
$\Rightarrow \frac{n}{2}(197-3n) < 0 \Rightarrow n > 65.\dot{6}$ | dM1 | Dependent on previous M. Solving quadratic in $n$. Accept $n=, n>, n<$.
$\Rightarrow n = 66$ | A1 | $n=66$ cso
**Alt I:**
$\sum_{r=1}^{n}(100-3r)<0 \Rightarrow \sum_{r=1}^{n}3r > \sum_{r=1}^{n}100$ | — | —
$\Rightarrow 3\frac{n(n+1)}{2} > 100n$ | M1 M1A1 | M1: Attempting to solve using $\sum r = \frac{n(n+1)}{2}$ and $\sum 100 = 100n$.
$\Rightarrow n > 65.6 \Rightarrow n = 66$ | — | —
---5. (i)
$$U _ { n + 1 } = \frac { U _ { n } } { U _ { n } - 3 } , \quad n \geqslant 1$$
Given $U _ { 1 } = 4$, find
\begin{enumerate}[label=(\alph*)]
\item $U _ { 2 }$
\item $\sum _ { n = 1 } ^ { 100 } U _ { n }$\\
(ii) Given
$$\sum _ { r = 1 } ^ { n } ( 100 - 3 r ) < 0$$
find the least value of the positive integer $n$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q5 [6]}}