Question

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$$\begin{cases} c(1) = \\dfrac{3}{16} \\\\ c(n) = c(n - 1) \\cdot 4 \\end{cases}$$

what is the $3^{\text{rd}}$ term in the sequence?
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Explanation:

Step1: Find the 2nd term

To find the 2nd term (\(c(2)\)), we use the recursive formula \(c(n)=c(n - 1)\cdot4\) with \(n = 2\) and \(c(1)=\frac{3}{16}\).
So, \(c(2)=c(2 - 1)\cdot4=c(1)\cdot4=\frac{3}{16}\cdot4=\frac{3}{4}\)

Step2: Find the 3rd term

To find the 3rd term (\(c(3)\)), we use the recursive formula \(c(n)=c(n - 1)\cdot4\) with \(n = 3\) and \(c(2)=\frac{3}{4}\).
So, \(c(3)=c(3 - 1)\cdot4=c(2)\cdot4=\frac{3}{4}\cdot4 = 3\)

Answer:

\(3\)