| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Limit of ratio of sums |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring telescoping series technique and limit analysis, but follows a highly structured approach with the identity provided. The telescoping sum is mechanical once recognized, and the limit analysis is straightforward application of dominant terms. More challenging than standard A-level due to the FP1 context and multi-step nature, but the scaffolding makes it accessible. |
| Spec | 4.06a Summation formulae: sum of r, r^2, r^34.06b Method of differences: telescoping series |
| Answer | Marks |
|---|---|
| \((N + 1/2)^6 - 1/64 = 6S_N + (5/4)N^2(N+1)^2 + 3N(N+1)/16\) | M1A1A1 |
| \(S_N = (1/6)(N+2)^5 - (5/24)N^2(N+1)^2 - (1/32)N(N+1) - 1/384\) | |
| Or \(\frac{1}{6}\left[(N+\frac{1}{2})^6 - (\frac{1}{2})^6 - \frac{5N^2(N+1)^2}{4} - \frac{3}{16}N(N+1)\right]\) | A1 |
| [4] | |
| (i) For \(\lambda = 6, S_x = 1/6\) | B2 |
| (ii) For \(\lambda > 6, S_x = 0\) | B1 |
| [3] |
$(N + 1/2)^6 - 1/64 = 6S_N + (5/4)N^2(N+1)^2 + 3N(N+1)/16$ | M1A1A1 |
$S_N = (1/6)(N+2)^5 - (5/24)N^2(N+1)^2 - (1/32)N(N+1) - 1/384$ |
Or $\frac{1}{6}\left[(N+\frac{1}{2})^6 - (\frac{1}{2})^6 - \frac{5N^2(N+1)^2}{4} - \frac{3}{16}N(N+1)\right]$ | A1 |
| [4] |
**(i)** For $\lambda = 6, S_x = 1/6$ | B2 |
**(ii)** For $\lambda > 6, S_x = 0$ | B1 |
| [3] |4 The sum $S _ { N }$ is defined by $S _ { N } = \sum _ { n = 1 } ^ { N } n ^ { 5 }$. Using the identity
$$\left( n + \frac { 1 } { 2 } \right) ^ { 6 } - \left( n - \frac { 1 } { 2 } \right) ^ { 6 } \equiv 6 n ^ { 5 } + 5 n ^ { 3 } + \frac { 3 } { 8 } n$$
find $S _ { N }$ in terms of $N$. [You need not simplify your result.]
Hence find $\lim _ { N \rightarrow \infty } N ^ { - \lambda } S _ { N }$, for each of the two cases\\
(i) $\lambda = 6$,\\
(ii) $\lambda > 6$.
\hfill \mbox{\textit{CAIE FP1 2010 Q4 [7]}}