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.

 

[tacman]

Yes, there is a short and appropriate method to solve this problem. We can use the binomial theorem, but instead of expanding the entire expression, we can use a shortcut.

First, let's rewrite the expression as (2^(1/2))^10 - 1^10. We can then apply the binomial theorem to (2^(1/2))^10, which will give us (10 choose 0)(2^(1/2))^10 - (10 choose 1)(2^(1/2))^9 + (10 choose 2)(2^(1/2))^8 - ... + (10 choose 9)(2^(1/2))^1 - (10 choose 10)(2^(1/2))^0.

Simplifying this, we get 2^5 - 10(2^(1/2))^9 + 45(2^(1/2))^8 - ... + 45(2^(1/2))^1 - 1.

Now, we can see that the terms with an even power of 2^(1/2) will cancel out, leaving us with only the terms with an odd power. This gives us (2^5 - 10(2^4) + 45(2^3) - ... + 45(2^1) - 1)(2^(1/2))^10.

Simplifying further, we get (32 - 160 + 360 - ... + 45 - 1)(2^(1/2))^10.

Finally, we are left with (2^(1/2) - 1)^10, which is in the desired form of k^(1/2) - (k-1)^(1/2) where k = 2.

Therefore, the expression (2^(1/2) - 1)^10 can be expressed as 2^(1/2) - (2-1)^(1/2), or simply 2^(1/2) - 1. This is a much shorter and more appropriate method compared to expanding the entire expression using the binomial theorem.