Solveeit Logo
Login

Question

Question: Consider a series of number defined as following x<sub>0</sub> = \(\sqrt{a}\), x<sub>1</sub> = \(\s...

Consider a series of number defined as following

x0 = a\sqrt{a}, x1 = a+a,\sqrt{a + \sqrt{a},}x2 = a+a+a,\sqrt{a + \sqrt{a + \sqrt{a}}}, ……. Where

a > 0 then lim⁡n→∞\lim_{n \rightarrow \infty} xn =

A

a\sqrt{a}

B

1+4a+12\frac{\sqrt{1 + 4a} + 1}{2}

C

1+4a+12a\frac{\sqrt{1 + 4a} + 1}{2a}

D

Can not be found

Answer

1+4a+12\frac{\sqrt{1 + 4a} + 1}{2}

Explanation

Solution

We have xn2x _ { n } ^ { 2 } = a + xn–1

It is easy to see that the variable xn increases. Let us show that all its values remain less than some constant number we have xn−12x _ { n - 1 } ^ { 2 } – xn–1 – a < 0 (Q xn–1 < xn)

Hence, < 0

Since the expression in the second bracket is positive, so we have xn–1 < (4a+1)+12\frac { \sqrt { ( 4 a + 1 ) } + 1 } { 2 }

Put xn–1 = xn = α  

From the original relation between xn and xn–1, we get

α2 – α – a = 0, α = and since α ≥ 0 we have

α = (4a+1)+12\frac { \sqrt { ( 4 a + 1 ) } + 1 } { 2 }.