Is there a quick formula to find the ratios; $\displaystyle\frac{r(k)}{rtotal}$ for $k=1 \dots n$ without calculating the numerator and the denominator where; 


$\displaystyle r(k) = \sum\limits\_{i=1}^{n} {{n^2 \choose (k-1)\*n+i}}$ and $\displaystyle rtotal = \sum\limits\_{k=0}^{n} {r(k)}$.


I know the denominator $\displaystyle rtotal={n^2 \choose 0} + {n^2 \choose 1} + \dots + {n^2 \choose n^2} = 2^{n^2}$


If I can find a formula for each numerator;


$\displaystyle r(1)={n^2 \choose 1} + \dots + {n^2 \choose n} = ?$


$\displaystyle r(2)={n^2 \choose n+1} + \dots + {n^2 \choose 2n} = ?$


$\dots$


$\displaystyle r(n)={n^2 \choose (n-1)\*n+1} + \dots + {n^2 \choose n^2} = ?$


then some cancelling would be possible yielding to easier calculations. I need the ratios for $1 < n < 100$ but for $n>40$, $rtotal$ gets huge in my program. 



Sum of combinations
If you are interested in approximations to your ratios, you may find the accepted answer (and some comments of mine) to this MathOverflow post useful: [Sum of 'the first k' binomial coefficients for fixed n](https://mathoverflow.net/questions/17202/sum-of-the-first-k-binomial-coefficients-for-fixed-n)  .


Essentially, the approximations are geometric series starting with the dominant term in your sums, and except for k near n/2, should serve you well, especially if you collapse adjacent terms in the form of n^2 +1 choose j. For k near n/2, Michael Lugo has some suggestions in his answer to the question above. If you do a search on "binomial coefficients sum", you will find other MathOverflow posts considering sums similar to yours which might help you.


I think getting a nice closed form for the exact values for your numerators in general is unsolvable.


Gerhard "Ask Me About System Design" Paseman, 2011.10.28


