Consider an = 2^(n-1)
a1, a2, a3. a4, … = 1, 10, 100, 100, ,,,
Use induction
3 a1 = 11
= a1 XOR 3 a1 = 01 XOR 11 = 10 = a2
now consider
an XOR 3an
= a^(n-1) XOR (2 an + 1 an )
= 0 2*n + 1 2^(n-1) XOR 1 2^n + 1 2^(n-1)
(done pair-wise)
= 1 2*n + 0 2^(n-1)
= 2^n = a_(n+1)
QED
i.e.: 01000… XOR 11000… = 10000…
Edited on October 31, 2018, 9:56 am