Explanation
Step-1: Expanding the terms
in+in+1+in+2+in+3
=in+in.i+in.i2+in.i3
Step-2: Taking in common
=in(1+i+i2+i3)
=in(1+i−1−i) [∵i2=−1,i3=−i]
=in(0)=0
Hence the correct option is (C) 0
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