View Full Version : a difficult one
viren_t2005
Dec3-05, 12:19 AM
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?
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.
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