Question

$\sqrt{2 + \sqrt{2 + \sqrt{2} } } + ........ \infty$
can you solve this question in imedetally?​

Answer 1

first let assume that

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2.......... \infty } } } } } = x$

Leaving first term remaining terms will also be = x .

So,

We get

$\sqrt{2+x} = x$

S.B.S

$2 + x = {x}^{2} \\ {x}^{2} - x - 2 = 0 \\ x(x-2)+1(x-2)= 0 \\ x = -1 \: \: \: \: \: \: or \: \: \: \: \: \: \: x = 2$

As,

$\sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2.......... \infty } } } } } = x$

And x = -1,2.

So,

$\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2.......... \infty } } } } } = 2$

or

$\sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2+ \sqrt{2.......... \infty } } } } } = -1$

But,

We know that sum of positive numbers will not be negative ,

So ,

$\sqrt{2 + \sqrt{2+ \sqrt{2 + \sqrt{2+ \sqrt{2.......... \infty } } } } } = 2$

Answer 2

Question:

$\large{\sqrt{2 + \sqrt{2 + \sqrt{2 \sqrt{2.....\infty=?}}}}}$

Answer: $\implies$ $\large{\boxed{\boxed{2}}}$

Explanation:

$\star$ Hey this can be solved by assuming the whole as A except the first being as it

$\implies$ $\large{\sqrt{2+A=x}}$

Squaring both the sides

$\implies$ 2 + A = A²

$\implies$ A²-A-2 = 0

$\implies$ (A-2)(A+1) =0

Here we get two cases

As A = +2 and A= -1

✰ Case 1

$\large{\sqrt{2 + \sqrt{2 + \sqrt{2 \sqrt{2.....\infty=+2}}}}}$

✰ Case 2

$\large{\sqrt{2 + \sqrt{2 + \sqrt{2 \sqrt{2.....\infty=-1}}}}}$

Sum of positive integers is positive

so Case 1 will be taken as the answer