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