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3 solutions
Has anyone found a method for reversing infinitely nested square roots? I've seen many solutions where an expression is shown to be equivalent to the question, but I've never seen anyone show how you could get from a nested radical back to a simple expression.
Why x + 1 ?
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Mathematical intuition... You might need to see his previous problems.... They are improvisations of the same concept (yet they are too good)... :D ...Anyways what idea did you used to solve this problem?
Yeah, I'm also curious as to how he got x + 1 as the function he should use...
You may have a look at my solution which is for (x+a) instead of (x+1), thus is a better generalization of the presented nested radical. Thank You
We know ( x + a ) = x 2 + a 2 + 2 a x = a 2 + x ( x + 2 a ) = a 2 + x 4 a 2 + x ( x + 4 a ) = a 2 + x 4 a 2 + x 1 6 a 2 + x ( x + 8 a ) = a 2 + x 4 a 2 + x 1 6 a 2 + x 6 4 a 2 + x . . . Now, setting x=a=1 we get, 2 = 1 + 4 + 1 6 + 6 4 + 2 5 6 + . . . Try out these crazy nested radicals!
How is this? 4x+1=(4x+1)^2
2nd green expression in your solution
Something isn t right here. If I put x=1 in the 2nd red expresion, that doesn t make the the equation of the problem
He did inverse calculations, which cancel out. Rooted and squared it, and the number didn't change.
x + 1 = ( x + 1 ) 2 = x 2 + 2 x + 1 = x 2 + ( 2 x + 1 ) 2 = x 2 + 4 x 2 + 4 x + 1 = x 2 + 4 x 2 + ( 4 x + 1 ) 2 = x 2 + 4 x 2 + 1 6 x 2 + 8 x + 1 = ⋯ Put x=1 to get the desired result i.e 2 = 1 + 4 + 4 2 + 4 3 + 4 4 + ⋯