Progression (Sequence and Series)HardBloom L3

Question

Let $\left\{ a_{n} \right\}$ consists of positive numbers and for any positive integer $n,\frac{a_{n} + 2}{2} = \sqrt{2s_{n}}$, where $s_{n} = \sum_{i = 1}^{n}\mspace{2mu} a_{i}$. Then

Options

A.$a_{21} = 82$
B.$a_{12} = 48$
C.$a_{13} = 50$
D.$a_{14} = 54$

Solution

Sol. [Key Concept: Positive-term recurrence: bound the sequence by induction.]

$$\frac{a_{1} + 2}{2} = \sqrt{2a_{1}} \Rightarrow \left( a_{1} - 2 \right)^{2} = 0 \Rightarrow a_{1} = 2$$

$$\begin{matrix} \left( a_{n} + 2 \right)^{2} = 8s_{n} \\ \left( a_{n - 1} + 2 \right)^{2} = 8s_{n - 1} \end{matrix}$$

✔ (a) $a_{21} = 4(21) - 2 = 82$

✘ (b) $a_{12} = 46 \neq 48$

✔ (c) $a_{13} = 50$

✔ (d) $a_{14} = 54$

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