$$ (\sqrt{2}+1 )^n=\sqrt{m-1}+\sqrt{m}$$

Question

anonymous · Answer

$$\color{red}{\text{show that for constants m,n} (\sqrt{2}+1 )^n=\sqrt{m-1}+\sqrt{m}}$$

experimentx · Answer

is m,n natural or real number?

anonymous · Answer

Let m and n be are positive integer

experimentx · Answer

it seems to be valid for any number http://www.wolframalpha.com/input/?i=solve+%281+%2B+sqrt%282%29%29%5Epi+%3D+sqrt%28m+-+1%29+%2B+sqrt%28m%29

anonymous · Answer

yes i was supprised when i teseted 1,2,3... but the question says
for some positive integer m.

anonymous · Answer

oh sorry its not pi
its n

anonymous · Answer

http://www.wolframalpha.com/input/?i=solve+%281+%2B+sqrt%282%29%29%5En+%3D+sqrt%28m+-+1%29+%2B+sqrt%28m%29

experimentx · Answer

interestingly i found the n=1,2,3, ... the value of m is given by http://oeis.org/search?q=2%2C9%2C50%2C+289%2C+1682&sort=&language=english&go=Search

anonymous · Answer

oh exaclty this is true since any triangular number 
is
$$T=\frac{n^2-n}{2}\implies 2T=n^2-n \implies (n-1/2)^2=2T+1/4 \implies 2(n-1/2)=\sqrt{8T+1}$$
stuck!

anonymous · Answer

but yeah thats true

anonymous · Answer

the equations not very clear they cud use a LaTex

experimentx · Answer

All Boils down to proving that 
`Table[Sum[Binomial[2 n, 2 k - 1] 2^{k - 2}, {k, 1, n}] , {n, 1, 10}]`
is equal to 
`Table[Sqrt[m (m - 1)/2] , {m, {1, 2, 9, 50, 289, 1682, 9801, 57122}}]`

experimentx · Answer

$$ (\sqrt{2}+1)^n = a+ b \sqrt 2 \
(\sqrt{2}-1)^n = -a+ b \sqrt 2 \
1 =  2 b^2 -a^2 \
2b^2 = a^2 - 1 \
$$
Follows that 
$$ (\sqrt{2} + 1)^n = \sqrt{a^2} + \sqrt{a^2 + 1}$$

experimentx · Answer

$$ \sum_{k=1}^n \binom{2n}{2k-1} 2^{k-2} = \sqrt{\frac{m(m+1)}{2}}$$
is a consequence of the question. I doubt if it could be proved other way.