What is the answer to t(n)=t(sqrt(n))+2 ?

The answer to t(n)=t(sqrt(n))+2 is t=(2(n+sqrt(n)))/(n^2-n);\quad n\ne 1,n\ne 0

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t(n)=t(sqrt(n))+2

t (n) = t (n) + 2

t = \frac{2 ( n + n )}{n ^{2} - n}; n \neq = 1, n \neq = 0

Solution steps

t (n) = t (n) + 2

t (n) = n t + 2

Expand

t (n) : n t

n t = n t + 2

Move

n t

to the left side

n t - n t = 2

Factor

n t - n t : t (n - n)

t (n - n) = 2

Divide both sides by

n - n; n \neq = 1, n \neq = 0

t = \frac{2 ( n + n )}{n ^{2} - n}; n \neq = 1, n \neq = 0

f, 17 f = x^{3} + y^{3} + 3 x^{2} - 18 y^{2} + 81 y + 5

x, x + 6 y = 277 x - 3

x, 3 x + 6 y = 1 [y]

x, f = 3 x - 9 f (\frac{1}{3})

3.81 = - 2 g + (- 4.19)

What is the answer to t(n)=t(sqrt(n))+2 ?
The answer to t(n)=t(sqrt(n))+2 is t=(2(n+sqrt(n)))/(n^2-n);\quad n\ne 1,n\ne 0