Question

$x = \sqrt{1 + 2 + \sqrt{1 + 3 + \sqrt{1 + 4 + \sqrt{1 + ......} } } }$ ,x=?
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Answer 1

ANSWER:

Let us take a numerical value of 'x'

That is 3

We know that

$\small{ \rm{ 3 = \sqrt{9} = \sqrt{1 + 8} = \sqrt{ 1 + 2 \times {4}} }} \\ \small{ \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 + 2 {\sqrt{ \red{16}} }} = \sqrt{1 + 2 \sqrt{1 + 15 } } }} \\ \small{ \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 + 2 \sqrt{1 +3 \times 5 } } = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{ \red{25}} } } }} \\ \small{ \rm{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \times 6} } }}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \small \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + 4 \sqrt{ \red{36}...........} } } }$

Here we observe that at last we have a perfect square marked in red colour

NEED TO PROVE :

${ \sf{ \to \sqrt{1 +n \sqrt{ {(n + 2)}^{2} } } }}$

${ \sf{\to \sqrt{1 + n \sqrt{1 + (n + 1) \sqrt{ {(n + 3)}^{3} } } } }}$

Hence x = 3

MORE INFORMATION:

This was a nice question asked by Srinivasa Ramanujan in 1919.He was a great mathematician.This question was also known as Ramanujan’s Radical Brain teaser

Answer 2

x = 3

x = √9

x = √(1+8)

x = √(1+2*4)

x = √(1+2√16) { ∵ 4 = √16 }

x = √1+2√(1 + 15)

x = √1 + 2√(1 + 3*5)

x = √1 + 2√1 + 3√25 { ∵ 5 = √25 }

x = √1 + 2√1 + 3√(1 + 24)

x = √1 + 2√1 + 3√(1 + 4*6)

x = √1 + 2√1 + 3√(1 + 4√36) { ∵ 6 = √36 }

x = √1 + 2√1 + 3√1 + 4√(1+35)

x = √1 + 2√1 + 3√1 + 4√(1+ 5*7)

x = √1 + 2√1 + 3√1 + 4√1+5√49 __________ { ∵ 7 = √49 }

Or, with This we can Say That :-

General Term of The Equation is :-

→ √1 + n√(n+2)²

→ √ 1 + n√ 1 + (n+1)√(n+3)² _______________

So, Now This Series Goes on And Sum of series will be 3 upto Infinity .

it was Proved by Most Famous mathematician named Srinivasa Ramanujan . { The Man who knew More Than Infinity . }