In a bit wise XOR operation:
a b a^b
-----------
0 0 0
0 1 1
1 0 1
1 1 0
XOR sum refers to successive XOR operations on integers.
Suppose you have numbers from 1 to N and you have to find their XOR sum then for N = 6, XOR sum will be 1^2^3^4^5^6 = 7.
1 = 001, 2 = 010, 3 = 011, 4 = 100, 5 = 101, 6 = 110
1^2 = 1^2 = 001^010 = 011 = 3
(1^2)^3 = 3^3 = 011^011 = 000 = 0
(1^2^3)^4 = 0^4 = 000^100 = 100 = 4
(1^2^3^4)^5 = 4^5 = 100^101 = 001 = 1
(1^2^3^4^5)^6 = 1^6 = 001^110 = 111 = 7 --> XOR sum
Hope this will help.