Question
It is given that the sequence $\left\{ a_{n} \right\}$ satisfies $a_{1} = 0,a_{n + 1} = a_{n} + 1 + 2\sqrt{1 + a_{n}}$ for $n \in N$. Then
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Solution
Sol. [Key Concept: Use the given recurrence to derive closed-form / monotonicity properties of the sequence.]
$$a_{n + 1} + 1 = \left( \sqrt{1 + a_{n}} + 1 \right)^{2}$$
$$\Rightarrow \ \sqrt{1 + a_{n + 1}} = \sqrt{1 + a_{n}} + 1$$
$\Rightarrow \left\{ \sqrt{1 + a_{n}} \right\}$ is an A.P. with first term $\sqrt{1 + a_{1}} = 1$ and common difference 1.
$$\Rightarrow \ \sqrt{1 + a_{n}} = 1 + (n - 1)1 = n \Rightarrow a_{n} = n^{2} - 1$$
✔ (a) $a_{100}{= 100}^{2 - 1 = 9999}$
✔ (b) $a_{2001}{= 2001}^{2 - 1 = 4004000}$
✘ (c) $a_{2001} = 4004000 \neq 4002000$
✔ (d) $a_{19}{= 19}^{2 - 1 = 360}$
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