Date | May 2015 | Marks available | 2 | Reference code | 15M.1.hl.TZ2.13 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Show that | Question number | 13 | Adapted from | N/A |
Question
Show that \(\frac{1}{{\sqrt n + \sqrt {n + 1} }} = \sqrt {n + 1} - \sqrt n \) where \(n \ge 0,{\text{ }}n \in \mathbb{Z}\).
Hence show that \(\sqrt 2 - 1 < \frac{1}{{\sqrt 2 }}\).
Prove, by mathematical induction, that \(\sum\limits_{r = 1}^{r = n} {\frac{1}{{\sqrt r }} > \sqrt n } \) for \(n \ge 2,{\text{ }}n \in \mathbb{Z}\).
Markscheme
\(\frac{1}{{\sqrt n + \sqrt {n + 1} }} = \frac{1}{{\sqrt n + \sqrt {n + 1} }} \times \frac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt {n + 1} - \sqrt n }}\) M1
\( = \frac{{\sqrt {n + 1} - \sqrt n }}{{(n + 1) - n}}\) A1
\( = \sqrt {n + 1} - \sqrt n \) AG
[2 marks]
\(\sqrt 2 - 1 = \frac{1}{{\sqrt 2 + \sqrt 1 }}\) A2
\( < \frac{1}{{\sqrt 2 }}\) AG
[2 marks]
consider the case \(n = 2\): required to prove that \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) M1
from part (b) \(\frac{1}{{\sqrt 2 }} > \sqrt 2 - 1\)
hence \(1 + \frac{1}{{\sqrt 2 }} > \sqrt 2 \) is true for \(n = 2\) A1
now assume true for \(n = k:\sum\limits_{r = 1}^{r = k} {\frac{1}{{\sqrt r }} > \sqrt k } \) M1
\(\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} > \sqrt k \)
attempt to prove true for \(n = k + 1:\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \) (M1)
from assumption, we have that \(\frac{1}{{\sqrt 1 }} + \ldots + \frac{{\sqrt 1 }}{{\sqrt k }} + \frac{1}{{\sqrt {k + 1} }} > \sqrt k + \frac{1}{{\sqrt {k + 1} }}\) M1
so attempt to show that \(\sqrt k + \frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} \) (M1)
EITHER
\(\frac{1}{{\sqrt {k + 1} }} > \sqrt {k + 1} - \sqrt k \) A1
\(\frac{1}{{\sqrt {k + 1} }} > \frac{1}{{\sqrt k + \sqrt {k + 1} }}\), (from part a), which is true A1
OR
\(\sqrt k + \frac{1}{{\sqrt {k + 1} }} = \frac{{\sqrt {k + 1} \sqrt k + 1}}{{\sqrt {k + 1} }}\) A1
\( > \frac{{\sqrt k \sqrt k + 1}}{{\sqrt {k + 1} }} = \sqrt {k + 1} \) A1
THEN
so true for \(n = 2\) and \(n = k\) true \( \Rightarrow n = k + 1\) true. Hence true for all \(n \ge 2\) R1
Note: Award R1 only if all previous M marks have been awarded.
[9 marks]
Total [13 marks]