| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Use standard formulae to show result |
| Difficulty | Moderate -0.3 This is a straightforward proof by induction question requiring standard techniques. Part (i) proves the sum of first n odd numbers equals n² using routine induction steps, while part (ii) applies this result with basic algebraic manipulation. Both parts are mechanical applications of well-known formulae with no novel insight required, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs4.06a Summation formulae: sum of r, r^2, r^3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n}(2r-1) = 2\sum_{r=1}^{n}r - n\) | M1 | Attempt to split into two sums (may be implied) |
| \(= n(n+1) - n = n^2\) | M1 A1 | Use of standard result for \(\sum r\), cao (must be in terms of n). SC Induction: B1 case \(n=1\); E1 sum to \(k+1\) terms correctly found; E1 argument completely correct |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{\displaystyle\sum_{r=1}^{n}(2r-1)}{\displaystyle\sum_{r=n+1}^{2n}(2r-1)} = \dfrac{n^2}{(2n)^2 - n^2}\) | M1 | Use of result from (i) in numerator of a fraction |
| M1 | Expressing denominator as \(\sum_{r=1}^{2n}\ldots - \sum_{r=1}^{n}\ldots\), need not be explicit, or other valid method | |
| A1 | Correct sums | |
| \(= \dfrac{n^2}{3n^2} = \dfrac{1}{3} = k\) | A1 | \(k = \dfrac{1}{3}\) |
| [4] |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n}(2r-1) = 2\sum_{r=1}^{n}r - n$ | M1 | Attempt to split into two sums (may be implied) |
| $= n(n+1) - n = n^2$ | M1 A1 | Use of standard result for $\sum r$, cao (must be in terms of n). SC Induction: B1 case $n=1$; E1 sum to $k+1$ terms correctly found; E1 argument completely correct |
| | **[3]** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{\displaystyle\sum_{r=1}^{n}(2r-1)}{\displaystyle\sum_{r=n+1}^{2n}(2r-1)} = \dfrac{n^2}{(2n)^2 - n^2}$ | M1 | Use of result from (i) in numerator of a fraction |
| | M1 | Expressing denominator as $\sum_{r=1}^{2n}\ldots - \sum_{r=1}^{n}\ldots$, need not be explicit, or other valid method |
| | A1 | Correct sums |
| $= \dfrac{n^2}{3n^2} = \dfrac{1}{3} = k$ | A1 | $k = \dfrac{1}{3}$ |
| | **[4]** | |
---5 (i) Show that $\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( 2 \mathrm { r } - 1 ) = \mathrm { n } ^ { 2 }$.\\
(ii) Show that $\frac { \sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( 2 \mathrm { r } - 1 ) } { \sum _ { \mathrm { r } = \mathrm { n } + 1 } ^ { 2 \mathrm { n } } ( 2 \mathrm { r } - 1 ) } = \mathrm { k }$, where $k$ is a constant to be determined.
\hfill \mbox{\textit{OCR MEI FP1 2015 Q5 [7]}}