| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation formula |
| Difficulty | Standard +0.3 This is a standard Further Maths induction question with two approaches. Part (i) requires algebraic manipulation of known formulae (expanding and substituting), while part (ii) follows the routine induction structure. Both parts are methodical rather than insightful, making this slightly easier than average for FP1 material, though still requiring careful algebra. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} r^2(r+1) = \sum_{r=1}^{n} r^3 + \sum_{r=1}^{n} r^2\) | M1 | Separation of sums (may be implied) |
| \(= \frac{1}{4}n^2(n+1)^2 + \frac{1}{6}n(n+1)(2n+1)\) | B1, M1 | One mark for both parts; Attempt to factorise (at least two linear algebraic factors) |
| \(= \frac{1}{12}n(n+1)\left[3n(n+1)+2(2n+1)\right]\) | ||
| \(= \frac{1}{12}n(n+1)(3n^2+7n+2)\) | A1 | Correct |
| \(= \frac{1}{12}n(n+1)(n+2)(3n+1)\) | E1 [5] | Complete, convincing argument |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sum_{r=1}^{n} r^2(r+1) = \frac{1}{12}n(n+1)(n+2)(3n+1)\) | ||
| \(n=1\): LHS \(=\) RHS \(= 2\) | B1 | 2 must be seen |
| Assume true for \(n = k\): \(\sum_{r=1}^{k} r^2(r+1) = \frac{1}{12}k(k+1)(k+2)(3k+1)\) | E1 | Assuming true for \(k\) |
| \(\sum_{r=1}^{k+1} r^2(r+1) = \frac{1}{12}k(k+1)(k+2)(3k+1) + (k+1)^2(k+2)\) | B1, M1 | \((k+1)\)th term; Attempt to factorise |
| \(= \frac{1}{12}(k+1)(k+2)\left[k(3k+1)+12(k+1)\right]\) | ||
| \(= \frac{1}{12}(k+1)(k+2)(3k^2+13k+12)\) | A1 | Correct |
| \(= \frac{1}{12}(k+1)(k+2)(k+3)(3k+4)\) | A1 | Complete convincing argument |
| \(= \frac{1}{12}((k+1)+1)((k+1)+2)(3(k+1)+1)\) | E1 | Dependent on previous A1 and previous E1 |
| But this is the given result with \(k+1\) replacing \(k\). Therefore if true for \(k\) it is true for \(k+1\). Since true for \(k=1\), true for \(k=1,2,3\) and so true for all positive integers. | E1 [8] | Dependent on first B1 and previous E1 |
# Question 10:
## Question 10(i):
| $\sum_{r=1}^{n} r^2(r+1) = \sum_{r=1}^{n} r^3 + \sum_{r=1}^{n} r^2$ | M1 | Separation of sums (may be implied) |
| $= \frac{1}{4}n^2(n+1)^2 + \frac{1}{6}n(n+1)(2n+1)$ | B1, M1 | One mark for both parts; Attempt to factorise (at least two linear algebraic factors) |
| $= \frac{1}{12}n(n+1)\left[3n(n+1)+2(2n+1)\right]$ | | |
| $= \frac{1}{12}n(n+1)(3n^2+7n+2)$ | A1 | Correct |
| $= \frac{1}{12}n(n+1)(n+2)(3n+1)$ | E1 [5] | Complete, convincing argument |
## Question 10(ii):
| $\sum_{r=1}^{n} r^2(r+1) = \frac{1}{12}n(n+1)(n+2)(3n+1)$ | | |
| $n=1$: LHS $=$ RHS $= 2$ | B1 | 2 must be seen |
| Assume true for $n = k$: $\sum_{r=1}^{k} r^2(r+1) = \frac{1}{12}k(k+1)(k+2)(3k+1)$ | E1 | Assuming true for $k$ |
| $\sum_{r=1}^{k+1} r^2(r+1) = \frac{1}{12}k(k+1)(k+2)(3k+1) + (k+1)^2(k+2)$ | B1, M1 | $(k+1)$th term; Attempt to factorise |
| $= \frac{1}{12}(k+1)(k+2)\left[k(3k+1)+12(k+1)\right]$ | | |
| $= \frac{1}{12}(k+1)(k+2)(3k^2+13k+12)$ | A1 | Correct |
| $= \frac{1}{12}(k+1)(k+2)(k+3)(3k+4)$ | A1 | Complete convincing argument |
| $= \frac{1}{12}((k+1)+1)((k+1)+2)(3(k+1)+1)$ | E1 | Dependent on previous A1 and previous E1 |
| But this is the given result with $k+1$ replacing $k$. Therefore if true for $k$ it is true for $k+1$. Since true for $k=1$, true for $k=1,2,3$ and so true for all positive integers. | E1 [8] | Dependent on first B1 and previous E1 |10 (i) Using the standard formulae for $\sum _ { r = 1 } ^ { n } r ^ { 2 }$ and $\sum _ { r = 1 } ^ { n } r ^ { 3 }$, prove that
$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + 1 )$$
(ii) Prove the same result by mathematical induction.
\hfill \mbox{\textit{OCR MEI FP1 2008 Q10 [13]}}