Ramanujan style double nested expression
Simplify the following expression:
The pattern here is that the coefficients in front of the square root is the sequence 1,2,3..., Pattern(n) = n, starting at n = 1
The terms being added to the square roots are 5,6... Pattern(n) = 4 + n, starting at n = 1
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SPOILER ALERT
Solution technique can be found: http://vixra.org/pdf/1310.0177v1.pdf
The answer is 3.
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1 solution
In case you're too lazy to read the pdf:
The principle is as follows:
( x + n ) 2 = x 2 + 2 x n + n 2
( x + n ) 2 = x 2 + n x + n ( x + n )
( x + n ) 2 = x 2 + n x − n e + n ( x + n + e )
( x + n ) 2 = x 2 + n ( x − e ) + n ( x + ( n + e ) )
x + n = x 2 + n ( x − e ) + n ( x + ( n + e ) )
From here it is easy to show then that for any value e of our choice we can make the substitution
x + n = x 2 + n ( x − e ) + n ( x + n = x 2 + ( n + e ) ( x − e ) + ( n + e ) ( x + ( n + 2 e ) ) )
Now we repeat this substitution indefinitely to get an infinitely nested expression.
The solution then requires us to notice that from the problem it is clear that
n = 1
Furthermore it is clear that e = 1
And therefore
x^2 + x - 1 = 5
Tells us that x = 2 is a viable candidate and therefore the solution is
(2 + 1) = 3