Infinitely nested radicals

x + n + a
= √(x + n + a)²
= √( x² + 2x(n + a) + (n + a)² )
= √( ax + (n + a)² + x² + 2nx + ax )
= √( ax + (n + a)² + x(x + 2n + a) )
= √( ax + (n + a)² + x(x+n + n + a) ) ... ①
Similarly, x+n + n + a = √( a(x+n) + (n + a)² + (x+n)(x+n+n + n + a) ) ,
Substitute it into ① :
x + n + a
= √( ax + (n + a)² + x√( a(x+n) + (n + a)² + (x+n)(x+n+n + n + a) ) )
= √( ax + (n + a)² + x√( a(x+n) + (n + a)² + (x+n)(x+2n + n + a) ) ) ... ②

Similarly, x+2n + n + a = √( a(x+2n) + (n + a)² + (x+2n)(x+2n+n + n + a) )
x+2n + n + a = √( a(x+2n) + (n + a)² + (x+2n)(x+3n + n + a) )
Substitute it into ② :
x + n + a
= √( ax + (n + a)² + x√( a(x+n) + (n + a)² + (x+n)√( a(x+2n) + (n + a)²
+ (x+2n)(x+3n + n + a) ) ) )
Similarly, x+3n + n + a = √( a(x+3n) + (n + a)² + (x+3n)√... )
∴ x + n + a
= √( ax + (n + a)² + x√( a(x+n) + (n + a)² + (x+n)√( a(x+2n) + (n + a)²
+ (x+2n)√( a(x+3n) + (n + a)² + (x+3n)√... ) ) ) )

Specially, put x = 2 , n = 1 , a = 0 :
2 + 1 + 0 = √( n² + 2√(n² + 3√(n² + 4√(n² + 5√(n² + ... ) ) ) ) )
3 = √( 1 + 2√(1 + 3√(1 + 4√(1 + 5√(1 + ... ) ) ) ) )