Python 2, 34 bytes

lambda n,k:(2*k-(k/n)*n)*-~(k/n)/2

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###A different approach:

I did not want to port my Ruby answer to Python (I knew somebody else would), so I started looking for a general formula:

If we call w=k/n then we can decompose the sum into the sum of integers from 1 to n: w*(w+1)/2 repeated n times and the rest (w+1)*(k-w*n). Add the two terms and simplify: we have the formula!

Python 2, 34 28 bytes

lambda n,k:(k+k%n)*(k/n+1)/2

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Thanks Martin Ender, Neil and Mr Xcoder for helping.