I am looking to check whether the hypersurface in $A^{n}$ defined by $x_{1}^{2} + x_2^{2} + .... + x_n^{2} = 0$ is a normal variety.....In general, are there any nice sufficiency conditions to prove normality?

Dear anonymous, Here is an expansion of what Georges said in the comment. I will assume, as you wrote, that you are a beginner in AG but not in math. And please do not feel too bad about diverietti's comment, for this site to function well we do need to keep a certain standard. That's why it is a good idea to use your real name and state your background. People here would be a lot more accommodating if they know exactly where you come from. As Georges wrote, normality is equivalent to two technical conditions: $R_1$ and $S_2$. $R_1$ means ``regular in codimension one". In the case of your interest, which is a hypersurface $f \in \mathbb A^n$, it can be checked easily (I will assume you work over $\mathbb C$). Just take the ideal $J$ generated by all the partial derivatives of $f$ and let $d$ be the dimension of $\mathbb C[x_1,\cdots, x_n]/J$. As long as $nd1\geq 2$, your hypersurface will be $R_1$. In your particular case, $J = (x_1,\cdots, x_n)$ and $d=0$, so as long as $n\geq 3$ you will be OK. But this procedures works for any hypersurface, for example $x^3+y^5+z^7$. The second condition $S_2$ is also known as ``Serre condition $S_2$". It is more technical to explain, and can actually be hard to check in general, but in this case, you are again in luck. Any hypersurface in $\mathbb A^n$ (for any $n$!) satisfies it. So, in summary, your quadric hypersurface is normal as long as $n\geq 3$, but hopefully what I wrote will be helpful in other cases you might be interested in. 


There is a simpler approach which works in this case, using the fact that this is a doublecover of affine $(n1)$space branched over a locus regular in codimension 1 (if the characteristic is not 2, as Karl Schwede pointed out). See Exercise 5.4.H (which gives a general useful tool) and Exercise 5.4.I(b) (which includes the question you ask) in the June 11, 2013 version of the notes available here. (Also, welcome to mathoverflow!) 


Given a CohenMacaulay affine ring $R$ over a perfect field, it satisfies $S_k$ for all $k$. Since $R$ is equidimensional, Jacobian criterion says that the singular locus is cut out by the Jacobian ideal $J$. It is then sufficient and necessary to have codim $J \ge 2$ in $R$ for $R$ to satisfy $R_1$. Assuming characteristic is not 2, the coordinate ring $S = k[x_1,\cdots,x_n]/(x_1^2+\cdots+ x_r^2)$ is cut out by a nonzerodivisor and is thus CohenMacaulay. Its singular locus is cut out by $I = (x_1,\cdots, x_r)$ which has codim $r1$ in $S$. Therefore $S$ is normal iff $r\ge 3$. 

