| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove summation with fractions |
| Difficulty | Standard +0.3 This is a standard proof by induction with a fractional summation formula. The base case is trivial (n=1 gives 1/3 = 1/3), and the inductive step requires algebraic manipulation of fractions but follows a predictable pattern. The partial fractions structure makes simplification straightforward. This is slightly easier than average because it's a textbook-style induction proof with no conceptual surprises, though the fraction manipulation requires care. |
| Spec | 4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(n=1\): \(\displaystyle\sum_{r=1}^{1}\frac{1}{(2r-1)(2r+1)} = \frac{1}{1\times3} = \frac{1}{3}\) and \(\dfrac{n}{2n+1} = \dfrac{1}{2\times1+1} = \dfrac{1}{3}\) (true for \(n=1\)) | B1 | Substitutes \(n=1\) into both sides; minimum expect \(\frac{1}{1\times3}\) and \(\frac{1}{2+1}\) |
| Assume general statement true for \(n=k\): \(\displaystyle\sum_{r=1}^{k}\frac{1}{(2r-1)(2r+1)} = \frac{k}{2k+1}\) is true | M1 | Assumes general result true for \(n=k\) |
| \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{(2r-1)(2r+1)} = \left``\frac{k}{2k+1}\right" + \frac{1}{(2k+1)(2k+3)}\) | M1 | Attempts to add \((k+1)\)th term to sum of \(k\) terms |
| \(= \dfrac{k(2k+3)+1}{(2k+1)(2k+3)}\) | dM1 | Combines fractions over correct common denominator \((2k+1)(2k+3)\); depends on previous M |
| \(= \dfrac{2k^2+3k+1}{(2k+1)(2k+3)} = \dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \dfrac{(k+1)}{2(k+1)+1}\) or \(\dfrac{k+1}{2k+3}\) | A1 | Correct algebraic work leading to \(\dfrac{(k+1)}{2(k+1)+1}\) or \(\dfrac{k+1}{2k+3}\) |
| As \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{(2r-1)(2r+1)} = \frac{(k+1)}{2(k+1)+1}\), true for \(n=k+1\). True for \(n=1\), true for \(n=k\) implies true for \(n=k+1\), so true for all \(n\in\mathbb{N}\) | A1cso | Correct induction statement with all three underlined elements; depends on all except B mark |
# Question 3:
| Working | Mark | Guidance |
|---------|------|----------|
| $n=1$: $\displaystyle\sum_{r=1}^{1}\frac{1}{(2r-1)(2r+1)} = \frac{1}{1\times3} = \frac{1}{3}$ and $\dfrac{n}{2n+1} = \dfrac{1}{2\times1+1} = \dfrac{1}{3}$ (true for $n=1$) | B1 | Substitutes $n=1$ into both sides; minimum expect $\frac{1}{1\times3}$ and $\frac{1}{2+1}$ |
| Assume general statement true for $n=k$: $\displaystyle\sum_{r=1}^{k}\frac{1}{(2r-1)(2r+1)} = \frac{k}{2k+1}$ is true | M1 | Assumes general result true for $n=k$ |
| $\displaystyle\sum_{r=1}^{k+1}\frac{1}{(2r-1)(2r+1)} = \left``\frac{k}{2k+1}\right" + \frac{1}{(2k+1)(2k+3)}$ | M1 | Attempts to add $(k+1)$th term to sum of $k$ terms |
| $= \dfrac{k(2k+3)+1}{(2k+1)(2k+3)}$ | dM1 | Combines fractions over correct common denominator $(2k+1)(2k+3)$; depends on previous M |
| $= \dfrac{2k^2+3k+1}{(2k+1)(2k+3)} = \dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \dfrac{(k+1)}{2(k+1)+1}$ or $\dfrac{k+1}{2k+3}$ | A1 | Correct algebraic work leading to $\dfrac{(k+1)}{2(k+1)+1}$ or $\dfrac{k+1}{2k+3}$ |
| As $\displaystyle\sum_{r=1}^{k+1}\frac{1}{(2r-1)(2r+1)} = \frac{(k+1)}{2(k+1)+1}$, true for $n=k+1$. True for $n=1$, true for $n=k$ implies true for $n=k+1$, so true for all $n\in\mathbb{N}$ | A1cso | Correct induction statement with all three underlined elements; depends on all except B mark |
---\begin{enumerate}
\item Prove by mathematical induction that, for $n \in \mathbb { N }$
\end{enumerate}
$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) } = \frac { n } { 2 n + 1 }$$
\hfill \mbox{\textit{Edexcel CP AS 2019 Q3 [6]}}