Number theory question on finite simple continued fractions I have to show that for any positive integer n, the following holds, some help: 
a) √(n^2+1) = [n,(2n) ̅]
b) √(n^2+2) = [n,(n,2n) ̅]
c) √(n^2+2n) = [n,(1,2n) ̅]
 A: Calculate the first few terms, as if you were working with a particular number. We do the first example, and leave the others to you.
Think about how one calculates the continued fraction for $\sqrt{n^2+1}$, with a particular $n$ like $10$. First we would take the greatest integer $\le \sqrt{n^2+1}$. This is $n$, and is our $a_0$.
Then we would subtract $n$, and find the reciprocal. the reciprocal of $\sqrt{n^2+1}-n$ is $\frac{1}{\sqrt{n^2+1}-n}$. Multiply top and bottom by $\sqrt{n^2+1}+n$. 
We get $\sqrt{n^2+1}+n$.  Now we need to take the greatest integer $\le \sqrt{n^2+1}+n$. This will be our $a_1$, and it is equal to $2n$.
Now we subtract $2n$ from $\sqrt{n^2+1}+n$, and take the reciprocal. We get $\frac{1}{\sqrt{n^2+1}-n}$, which is $\sqrt{n^2+1}+n$. We take the greatest integer in that, which is $2n$, so $a_2=2n$. and now when we subtract $2n$ and flip, again we get $\sqrt{n^2+1}+n$, so $a_3=2n$. and so on forever: The continued fraction is $\left<n:2n,2n,2n,\dots\right>$.  
Remark: It will be worthwhile, in this case and the others, to take a particular $n$, and compute, using a calculator. You will observe a certain pattern, and a little work will show that the pattern must always hold. 
