A) \[l{{+}^{n}}Z\]
B) \[l+({{2}^{n}}-1)Z\]
C) \[l-({{2}^{n}}-1)Z\]
D) None of these
Correct Answer: B
Solution :
Since, Z is idempotent, then \[{{Z}^{2}}=Z\] \[\Rightarrow \] \[{{Z}^{3}},\,{{Z}^{4}},...{{Z}^{n}}=Z\] ?(i) \[\therefore \] \[{{(I+Z)}^{n}}{{=}^{n}}{{C}_{0}}{{I}^{n}}+{{\,}^{n}}{{C}_{1}}{{I}^{n-1}}Z{{+}^{n}}{{C}_{2}}{{I}^{n-2}}\cdot {{Z}^{2}}\] \[...+{{\,}^{n}}{{C}_{n}}{{Z}^{n}}\] \[=I+\,{{(}^{n}}{{C}_{1}}Z{{+}^{n}}{{C}_{2}}Z+...{{+}^{n}}{{C}_{n}}Z)\] \[=I+{{(}^{n}}{{C}_{1}}{{+}^{n}}{{C}_{2}}+{{...}^{n}}{{C}_{2}})Z\] \[=I+({{2}^{n}}-1)Z\]
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