June 2025 Paper 3 Q11

Notes

M1: Finding common denominator or attempt to rationalise denominator for at least one fraction e.g. \(\frac{1}{\sqrt{n+1}-1}\cdot\frac{\sqrt{n+1}+1}{\sqrt{n+1}+1} - \frac{1}{\sqrt{n+1}+1}\cdot\frac{\sqrt{n+1}-1}{\sqrt{n+1}-1}\) or \(\frac{\sqrt{n+1}+1}{n} - \frac{\left(\sqrt{n+1}-1\right)}{n}\) oe
Condone consistently considering any general term in the sequence e.g. \(a_{k+1}\) for all marks \(\dfrac{1}{\sqrt{k+2}-1}\cdot\dfrac{\sqrt{k+2}+1}{\sqrt{k+2}+1} - \dfrac{1}{\sqrt{k+2}+1}\cdot\dfrac{\sqrt{k+2}-1}{\sqrt{k+2}-1}\)

A1: Correct simplification
\(a_{k+1}\) leads to \(\frac{2}{k+1}\)

A1: nfww
Correct explanation of why the general term must be rational e.g. in form \(\frac{p}{q}\) oe \(p\), \(q\) integers
Condone rational as \(n\) an integer or \(k\) integer
‘\(\frac{2}{n}\) is rational’ alone is A0 - must include comment that \(n\) is an integer

Notes

M1: Must relate to \(\frac{2}{n}\), \(a_n \to 0\) is not sufficient
Allow "\((n > 0\) and\()\ \lim\limits_{n\to\infty}\frac{2}{n} = 0\)"
Condone omission of \(n > 0\)
May find the difference between 2 terms or use inequalities
Allow \(\mathrm{f}(x) = \frac{2}{x} \Rightarrow \mathrm{f}^\prime(x) = -\frac{2}{x^2} < 0\) (condone \(n\) instead of \(x\))
Condone listing terms for M1 A0
FT their \(\frac{2}{n}\) provided in the form \(\frac{b}{n+c}\) where \(b, c\) integers

A1: See appendix
Convincing conclusion e.g. difference between terms is negative, so sequence is decreasing or \(\frac{2}{n+1} < \frac{2}{n}\), so decreasing or function \(\mathrm{f}(x)\) is decreasing therefore sequence \(a_n\) decreasing
A1 is only available if their \(\frac{2}{n}\) is correct for their term e.g. \(\dfrac{2}{k+1}\) from considering \(a_{k+1}\)
SC B1 for ‘decreasing’ following M0 for a correct but incomplete argument. But not just for ‘decreasing’ unsupported.

Appendix: exemplar responses for Q11(b)