Solving (2^1/2-1)^10: A Short & Appropriate Method?

[viren_t2005]

express (2^1/2-1)^10 in the form k^1/2-(k-1)^1/2 where k is a positive integer.{the square roots need not be irrational}we can do this by binomial theorem but it is very tedious.is there a short & appropriate method to solve this problem?

 

[Tide]

Just solve the equation

$$\sqrt{k} - \sqrt{k-1} = (\sqrt 2 - 1)^{10}$$

algebraically for k. I get k = 11,309,769. This will be a mess unless you try something like

$$\sqrt k - \sqrt {k-1} = N$$

from which

$$\sqrt k + \sqrt {k-1} = \frac {1}{N}$$

leading to

$$2\sqrt k = N + \frac {1}{N}$$

This is easy to solve for k and the solution can be simplified to what I showed above.