AS June 2019 Paper 1 Q3
3. Prove by mathematical induction that, for \(n \in \mathbb{N}\)
\[\sum_{r=1}^{n}\frac{1}{(2r - 1)(2r + 1)} = \frac{n}{2n + 1}\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(n = 1\), \(\displaystyle\sum_{r=1}^{1}\frac{1}{(2r - 1)(2r + 1)} = \frac{1}{1 \times 3} = \frac{1}{3}\) and \(\dfrac{n}{2n + 1} = \dfrac{1}{2 \times 1 + 1} = \dfrac{1}{3}\) (true for \(n = 1\)) | B1 | 2.2a |
| Assume general statement is true for \(n = k\). So assume \(\displaystyle\sum_{r=1}^{k}\frac{1}{(2r - 1)(2r + 1)} = \frac{k}{2k + 1}\) is true. | M1 | 2.4 |
| \(\displaystyle\left(\sum_{r=1}^{k+1}\frac{1}{(2r - 1)(2r + 1)} = \right)\ \text{“}\frac{k}{2k + 1}\text{”} + \frac{1}{(2k + 1)(2k + 3)}\) | M1 | 2.1 |
| \(= \dfrac{k(2k + 3) + 1}{(2k + 1)(2k + 3)}\) | dM1 | 1.1b |
| \(= \dfrac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \dfrac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \dfrac{(k + 1)}{2(k + 1) + 1}\) or \(\dfrac{k + 1}{2k + 3}\) | A1 | 1.1b |
| As \(\displaystyle\sum_{r=1}^{k+1}\frac{1}{(2r - 1)(2r + 1)} = \frac{(k + 1)}{2(k + 1) + 1}\) then the general result is true for \(n = k + 1\) As the general result has been shown to be true for \(n = 1\), and true for \(n = k\) implies true for \(n = k + 1\), so the result is true for all \(n \in \mathbb{N}\) | A1cso | 2.4 |
| (6) | ||
| (6 marks) |
Notes
B1: Substitutes \(n = 1\) into both sides of the statement to show they are equal. As a minimum expect to see \(\dfrac{1}{1 \times 3}\) and \(\dfrac{1}{2 + 1}\) for the substitutions. (No need to state true for \(n = 1\) for this mark.)
M1: Assumes (general result) true for \(n = k\). (Assume (true for) \(n = k\) is sufficient – note that this may be recovered in their conclusion if they say e.g. if true for \(n = k\) then ... etc.)
M1: Attempts to add \((k + 1)\)th term to their sum of \(k\) terms. Must be adding the \((k + 1)\)th term but allow slips with the sum.
dM1: Depends on previous M. Combines their two fractions over a correct common denominator for their fractions, which may be \((2k + 1)^2(2k + 3)\) (allow a slip in the numerator).
A1: Correct algebraic work leading to \(\dfrac{(k + 1)}{2(k + 1) + 1}\) or \(\dfrac{k + 1}{2k + 3}\)
A1: cso Depends on all except the B mark being scored (but must have an attempt to show the \(n = 1\) case). Demonstrates the expression is the correct for \(n = k + 1\) (both sides must have been seen somewhere) and gives a correct induction statement with all three underlined statements (or equivalents) seen at some stage during their solution (so true for \(n = 1\) may be seen at the start).
For demonstrating the correct expression, accept giving in the form \(\dfrac{(k + 1)}{2(k + 1) + 1}\), or reaching \(\dfrac{k + 1}{2k + 3}\) and stating “which is the correct form with \(n = k + 1\)” or similar – but some indication is needed.
Note: if mixed variables are used in working (\(r\)’s and \(k\)’s mixed up) then withhold the final A.
Note: If \(n\) is used throughout instead of \(k\) allow all marks if earned.