A sequence of positive numbers is defined by $$a_{n+1} = \sqrt{a_n^2 + 3},\ n > 1,$$ $$a_1 = 2$$ (a) Find $a_2$ and $a_3$, leaving your answers in surd form - Edexcel - A-Level Maths Pure - Question 7 - 2010 - Paper 1

Next, we find $a_3$ by substituting $n = 2$:

$a_3 = \sqrt{a_2^2 + 3} = \sqrt{(\sqrt{7})^2 + 3} = \sqrt{7 + 3} = \sqrt{10}.$

Therefore,

$a_3 = \sqrt{10}.$

To show that $a_5 = 4$, we need to compute the values from $a_4$ and $a_5$ systematically.

First, we calculate $a_4$ using $n = 3$:

$a_4 = \sqrt{a_3^2 + 3} = \sqrt{(\sqrt{10})^2 + 3} = \sqrt{10 + 3} = \sqrt{13}.$

Then we compute $a_5$:

$a_5 = \sqrt{a_4^2 + 3} = \sqrt{(\sqrt{13})^2 + 3} = \sqrt{13 + 3} = \sqrt{16} = 4.$

Thus, we have shown that:

$a_5 = 4.$