Answer:
The answer is "[tex]\bold{\frac{1}{4}<k\leq 2+\sqrt{3}}[/tex]"
Step-by-step explanation:
Given:
[tex]\ S_1 = k \\\\ S_{n+1} = \sqrt{4S_n -1}[/tex] [tex]_{where} \ \ n \geq 1[/tex]
In the above-given value, [tex]S_n[/tex] is required for the monotone decreasing so, [tex]S_2 :[/tex]
[tex]\to \sqrt{4k-1} \leq \ k=S_1\\\\[/tex]
square the above value:
[tex]\to k^2-4k+1 \leq 0\\\\\to k \leq 2+\sqrt{3} \ \ \ \ \ and \ \ 4k+1 >0\\\\[/tex]
[tex]\bold{\boxed{\frac{1}{4}<k\leq 2+\sqrt{3}}}[/tex]
